Strategy and Tactics in Problem Solving
"...small solved and unsolved problems lead to larger solved and unsolved problems which in turn lead to important mathematical results."
Murray S. Klamkin
"Each problem that I solved became a rule which served afterwards to solve other problems."
Descartes
Inroduction
Solving problems is an activity not unique to mathematics. There are strategies of problem solving that are applicable to solving problems in any human endeavor. But every field and every situation call for specific knowledge and specific habits of mind to apply to solving problems. I therefore distinguish the universal strategies from field specific tactics.
As a young assistant professor I was looking for a desk. I found one new plain table of suitable dimensions in a carpenter shop for a price that in my estimate would not even cover the cost of materials. To my pointing that out the owner replied, "That's OK. This is because I know how." 35 years later I still have that table, although it now plays a different role in our household.
There is nothing specific to mathematics in George Polya's four stages of problem solving:
- Understanding the problem
- Devising a plan
- Carrying out the plan
- Looking back
To these I can add considering special cases as a stepping stone in understanding the problem; seeking an analogous or related problem; splitting the problem into smaller ones (which may be also regarded as moving to Polya's second stage.) Generalization is the opposite of specialization and is often throws revealing light on the problem's true nature. These all are strategic tools of problem solving. However, details of implementation are, of necessity, field specific.
At the time of writing, this site contained about 5500 pages, at least one third of which is devoted to solving problems, many of olympiad caliber. It may take time to classify them as to the problem solving tactics employed, but I plan to do that in a systematic manner.
I found a detailed classification in an article by Murray Klamkin, cited below. I added a few, and for some have not yet come across suitable examples. I'll be looking to fill the gaps.
I would very much appreciate comments and additional examples.
There is always satisfaction in solving a problem. The most common way to approach a problem is by identifying a general class to which the problem belongs and using a method (if such exists) that is applicable to the problems of that class. Enjoyment of having a problem solved is more a result of detecting and exploiting the problem's peculiarities. As a rule, those peculiarities once observed define a refined class of problems which succumb to the same method of solution. This is actually a problem solver learning process: looking back at the just solved problem, make a note as to what features made the problem solvable by the method you employed.
One of the best problem solving strategies is do something; do not get (and stay) flustered if you do not see a solution to a problem right away. Try one of the tactics, force yourself to speak aloud - something will come out. Ray Bradbury, when short on ideas, taught himself to work with a dictionary picking random words and trying to put them together into something meaningful and relevant. So try.
Also, note that some of the tactics was previously mentioned - with additional advice - in a short assay of what constitutes a proof.
Further reading
I highly recommend the 1981 article The heuristic of George Polya and its relation to artificial intelligence by Alan Newell that offers penetrating analysis of Polya's problem solving heuristics prior to the discussion on its applicability to AI.
- R. Gelca, T. Andreescu, Putnam and Beyond, Springer, 2007
- A. Engel, Problem-Solving Strategies, Springer Verlag, 1998
- M. S. Klamkin, Mathematical Creativity in Problem Solving II, in In Eves' Circles, J. M. Anthony (ed.), MAA, 1994
- G. Polya, How to Solve It, Princeton University Press, 1973
- G. Polya, Mathematical Discovery, John Wiley & Sons, 1981
- G. Polya, Mathematics and Plausible Reasoning, v 1, Princeton University Press, 1954
- P. Zeitz, The Art and Craft of Problem Solving, John Wiley & Sons, 1999
Special cases
Specializaton, i.e., considering special cases, or introducing additional restrictions, while trying to solve a problem may highlight aspects of the original problem that have perhaps been missed at first glance. Success in solving special cases may be reassuring of the validity of the original claim, while failure to solve a special case may prompt the solver to look for a counterexample which will also serve to disprove the larger problem.
Specialization is opposite to generalization; it is wonderful that both are valuable tools in the arsenal of a problem solver.
Generalization
There is a whole page at the site devoted to G. Polya's thesis that quite often more general or more strict problems are easier to solve than the more special ones. The page explains the concept and features several lists of solved sample problems.
Analogy
The Pythagorean theorem has two distinct analogues in the $3D$ space. For the parallelipiped with sides $a,b,c$ and diagonal $d,$ we have $a^{2}+b^{2}+c^{2}=d^{2}.$ For a tetrahedron with three faces having right angles at the shared vertex and areas $A,B,C,$ $A^{2}+B^{2}+C^{2}=D^{2},$ where $D$ is the area of the remaining face.
Being analogous is very much close to being similar but the latter is usually more specific.
Related problems
Look back
Subproblems
Solve differently and compare
- Pythagorean Theorem
- Square root of 2 is irrational
- Infinitude of Primes
- Butterfly Theorem
- Concyclic Circumcenters: A Dynamic View
- Napoleon's Theorem
- Morley's Theorem
- Bottema's Theorem
- Another simple integral
- Middle School Problem from a Moscow Olympiad
- Breaking Chocolate Bars
- Four Travelers Problem
- Chameleons of Three Colors
- Fractions on a Binary Tree
- Countability of Rational Numbers
- All about altitudes
- Angles in 4-5-6 Triangle
- Probability of Two Integers Being Coprime
- Circle through the Incenter
Relaxation
Symmetry
Continuity
An argument by continuity assumes the presence of a continuous function whose properties could be used to solve a given problem. For example, the area can be assumed (and eventually proved) to be a continuous function of a geometric shape (Note, though, that not all planar shapes may be assigned a numeric area in any sensible way. The common ones - polygons, circles, ellipses - do have areas which depend continuously on their shape.)
Similarity
In mathematics, the word "similarity" may be encountered in several different contexts. There is a special page with explanations and a variety of examples: What Is Similarity?
WLOG
Without loss of generality - WLOG, for short - is a frequent stratagem in problem solving. It is incredibly powerful, although, at first sight, often appears innocuous. The essence of WLOG is in making a random selection among a number of available ones for the reason of all possible variants being equipotent, i.e., leading to exactly same procedure and result. For further explanation, examples and link, check a separate file.
Reductio ad absurdum
One of the methods of proof, is "reductio ad absurdum" - assuming what is to be proved false and getting a contradiction as a result.
Mathematical induction
Many many examples have been collected on a separate page.
Pigeonhole principle
Many many examples have been collected on a separate page.
Generating functions
Many many examples have been collected on a separate page.
Level curves
Linearity
Extremal Principle
Convexity
Inequalities
- Schur's Inequality $(x^t(x-y)(x-z)+y^t(y-z)(y-x)+z^t(z-x)(z-y)\ge 0)$
- An Application of Schur's Inequality $\small{\left(5\left(\sum\sqrt{x+y}\right)\left(\sum\sqrt{(x+y)(y+z)}\right)\ge\left(\sum\sqrt{x+y}\right)^3+18\sqrt{(x+y)(y+z)(z+x)}\right)}$
- Powers and Fractions Inequality $\left(\displaystyle\sum_{cycl}\frac{a^3b^3}{c^5}\ge \sum_{cycl}\frac{ab}{c}\right)$
- Three Junior Problems from Vietnam $\left(\left(\begin{array}{ccc}0&y&z\\x&0&z\\x&y&0\end{array}\right)\left(\begin{array}\;a\\b\\c\end{array}\right)=\left(\begin{array}\;x\\y\\z\end{array}\right)\right)$
- Newton's and Maclaurin's Inequalities $\left(E_k(X)\right)^2\gt E_{k-1}(X)E_{k+1}(X),$ $\left(\left(E_k(X)\right)^2\gt E_{k-1}(X)E_{k+1}(X)\right)$
- Rearrangement Inequality $\left(\displaystyle\sum_{i=1}^{n}x_iy_{n+1-i}\le\sum_{i=1}^{n}x_iy_{\sigma(i)}\le\sum_{i=1}^{n}x_iy_i\right)$
- Chebyshev Inequality $\left(\displaystyle n\sum_{i=1}^{n}x_iy_{i}\ge\sum_{i=1}^{n}x_i\sum_{i=1}^{n}y_i\right)$
- Chebyshev Holds a Key $\left(\displaystyle\sum_{k=1}^n\left[\left(\sum_{i=1}^k\sin a_i\right)\left(\sum_{i=1}^k\cos a_i\right)\right]\le\frac{n(n+1)(2n+1)}{12}\right)$
- Jensen's Inequality $\left( f(\lambda x_1+(1-\lambda )x_2)\le \lambda f(x_1)+(1-\lambda )f(x_2)\right)$
- Muirhead's Inequality $\left(\mathbf{a}\succ\mathbf{b}\;\right)$ implies $\left([\mathbf{a}]\ge [\mathbf{b}]\right)$
- Bergström's inequality $\left(\displaystyle \frac{x_1^2}{a_1}+\frac{x_2^2}{a_2}+\cdots+\frac{x_n^2}{a_n}\ge\frac{(x_1+x_2+\cdots+x_n)^2}{a_1+a_2+\cdots+a_n}\right)$
- Dorin Marghidanu's Example II $\left(\displaystyle \sum_{k=1}^n\frac{a^2_k}{a_{\sigma(k)}-1}\ge 4n\right)$
- Radon's Inequality and Applications $\left(\displaystyle \frac{x_1^{p+1}}{a_1^p}+\frac{x_2^{p+1}}{a_2^p}+\cdots+\frac{x_n^{p+1}}{a_n^p}\ge\frac{(x_1+x_2+\cdots+x_n)^{p+1}}{(a_1+a_2+\cdots+a_n)^p}\right)$
- Dorin Marghidanu's Example for Radon's Inequality $\left(\displaystyle\frac{a^{m+n+1}}{b^mc^n}+\frac{b^{m+n+1}}{c^ma^n}+\frac{c^{m+n+1}}{a^mb^n}\ge a+b+c\right)$
- Jordan and Kober Inequalities, PWW $\left(\displaystyle x\in\left(0,\frac{\pi}{2}\right)\,\Longrightarrow\,\sin x\gt\frac{2x}{\pi}\right)$
- A Mathematical Rabbit out of an Algebraic Hat $\left(x^{2} + y^{2} + z^{2} - xy - yz - zx \ge 0\right)$
- An Inequality With an Infinite Series $\left(\displaystyle \sum_{n=1}^{1000}\frac{1}{n^{3}+3n^{2}+2n} \lt \frac{1}{4}\right)$
- An Inequality: $\left(1/2 \cdot 3/4 \cdot 5/6 \cdot \ldots \cdot 99/100 \lt 1/10\right)$
- $\left(\text{A Low Bound for }1/2 \cdot 3/4 \cdot 5/6 \cdot \ldots \cdot (2n-1)/2n\right)$
- An Inequality: Easier to prove a subtler inequality
- Inequality with Logarithms (logπ3 + log3π > 2)
- An inequality: 1 + 1/4 + 1/9 + ... less than 2
- Inequality with Harmonic Differences $\left(\displaystyle\frac{2n}{3n+1} \lt \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{n+n}\lt \frac{25}{36}\right)$
- An Inequality by Uncommon Induction $\left(\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}\gt \frac{3n}{2n+1}\right)$
- Hlawka's Inequality $\left(|x+y|+|y+z|+|z+x|\le |x|+|y|+|z|+|x+y+z|\right)$
- An Application of Hlawka's Inequality $\left(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\le a+b+c+\sqrt{a^2+b^2+c^2-ab-bc-ca}\right)$
- An Application of Hlawka's Inequality II $\left(\displaystyle xPA+yPB+zPC+\frac{a\sqrt{3}}{2}PM\ge (y+z)PA'+(z+x)PB'+(x+y)PC'\right)$
- Geometric Proof of Hlawka's Inequality $\left(|\vec{a}|+|\vec{b}|+|\vec{c}|+|\vec{d}|\ge |\vec{a}+\vec{d}|+|\vec{b}+\vec{d}|+|\vec{c}+\vec{d}|\right)$
- Hlawka-like Dinca's Inequality $\left((x+y)^{\alpha}+(y+z)^{\alpha}+(z+x)^{\alpha}\le x^{\alpha}+y^{\alpha}+z^{\alpha}+(x+y+z)^{\alpha}\right)$
- Hlawka-like Inequality for Convex Functions $\left(g(x) + g(y) + g(z) + g(x + y + z) \ge g(x + y) + g(x + z) + g(y + z)\right)$
- Hlawka in Convex Quadrilateral $\left(AB+BC+CD+DA\gt AC+BD+2EF\right)$
- Quadrilateral Inequality
- Problem 4165 from Crux Mathematicorum $\left(\displaystyle \left|\sum_{i=1}^4x_i\right|+2\sum_{i=1}^4|x_i|\ge 6\sqrt[6]{\prod_{1\le i\lt j\le 4}|x_i+x_j|}\right)$
- An Inequality in Determinants $\left(\left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^4 & b^4 & c^4\\ \end{array}\right|\ge \left|\begin{array}{ccc} 1 & 1 & 1\\ a & b & c\\ a^3 & b^3 & c^3. \end{array}\right|\right)$
- Application of Cauchy-Schwarz Inequality $\left(\displaystyle\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\le\frac{ab+bc+ca}{2}\right)$
- An Inequality from Tibet $\left((bcde+acde+abde+abce+abcd)^4\ge 125(a+b+c+d+e)(abcde)^3\right)$
- An Inequality from Morocco $\left(\displaystyle 4|(a-b)(b-c)(c-d)(d-a)|\le abcd\right)$
- An Inequality for Mixed Means $\small{\left(\displaystyle \sqrt[m]{\prod_{j=1}^{m}\left[a_j+\frac{1}{n}\sum_{i=1}^{n}x_i\right]}\ge\frac{1}{n}\sum_{i=1}^{n}\sqrt[m]{\prod_{j=1}^{m}(a_j+x_i)} \ge\prod_{i=1}^{n}\prod_{j=1}^{m}\sqrt[mn]{a_j+x_i}\right)}$
- An Inequality in Integers $\left(\displaystyle\frac{a}{b}+\frac{1}{ab}\lt\sqrt{7}\right)$
- An Inequality in Integers II $\left(\displaystyle\frac{a^3+b^3+c^3}{3}-abc\ge 3k\right)$
- An Inequality in Integers III $\left(\displaystyle\sum_{i,j=1}^n \frac{i^2j^2}{i+j-1}\gt n^3\right)$
- An Inequality with Exponents $\left(\displaystyle e^{\frac{4}{e}}\left(b\cdot a^{2\sqrt{a}}+c\cdot b^{2\sqrt{b}}+a\cdot c^{2\sqrt{c}}\right)\ge 3\sqrt[3]{abc}\right)$
- Exponential Inequalities for Means $\left(\displaystyle A_n^{G_nH_n}\le G_n^{A_nH_h}\le H_n^{A_nG_n}\right)$
- A Simple Inequality in Three Variables $\left(\displaystyle\sum_{cyc}ac\left(\frac{1}{2a+b}+\frac{1}{2c+b}\right)\le \sqrt{3(a^2+b^2+c^2)}\right)$
- An Asymmetric Inequality $\left((a+b)(a+c)\ge 2\sqrt{abc(a+b+c)}\right)$
- Linear Algebra Tools for Proving Inequalities $\left(\displaystyle\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\ge 2.\right)$
- An Inequality with a Generic Proof $\left(2\sqrt{b}+\sqrt{aS+b}\le \sqrt{ax+b}+ \sqrt{ay+b}+ \sqrt{az+b}\le \sqrt{3aS+9b}\right)$
- A Generalization of an Inequality from a Romanian Olympiad $\left(\displaystyle \frac{2}{n-1}\sum_{1\le i\lt j\le n}a_{i}a_{j}\le\ln\left(\prod_{k=1}^{n}a_{k}^{a_{k}}\right)+\frac{1}{e}\sum_{k=1}^{n}e^{a_{k}}\right)$
- Area Inequality in Trapezoid $\left([PQRS]\le\frac{1}{4}[ABCD]\right)$
- Improving an Inequality $\left(\displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\gt 2(\sqrt{n+1}-1)\right)$
- RomanoNorwegian Inequality $\left(\sqrt{\cot A} + \sqrt{\cot B} + \sqrt{\cot C} \ge 2\right)$
- Inequality with Nested Radicals II $\left(\displaystyle\sqrt{n+1}\le\sqrt{n+\sqrt{(n-1)+\sqrt{\ldots+\sqrt{2+\sqrt{1}}}}}\le\sqrt{2n-1}\right)$
- Inequality with Powers And Radicals $\left(\displaystyle\sum_{cycl}\sqrt[6]{ab^2c^3}\ge\sum_{cycl}\sqrt[30]{a^9b^{10}c^{11}}\right)$
- Inequality with Two Minima $\left(\displaystyle\sqrt{a-c}+\sqrt{b-c}\le\min \left\{\sqrt{\frac{ab}{c}},\sqrt{2(a+b-2c)} \right\}\right)$
- Simple Inequality with Many Faces And Variables $\left(\displaystyle \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1}\ge 20\right)$
- An Inequality with Determinants $\left(\displaystyle\left|\begin{array}{cccc}\,a&-b&0&0\\0&b&-c&0\\0&0&c&-d\\1&1&1&1+d\end{array}\right|\ge 3\sqrt[3]{4}(abcd)^{5/6}\right)$
- An Inequality with Determinants II $\left(\displaystyle \Delta=\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ a & b & c & d\\ a^2 & b^2 & c^2 & d^2\\ \frac{1}{a^2} & \frac{1}{b^2} & \frac{1}{c^2} & \frac{1}{d^2} \end{array}\right|\lt \frac{1}{abcd}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\right)$
- An Inequality with Determinants III $\left(\displaystyle \left|\begin{array}{ccccc} \,0 & x^2 & y^2 & z^2 & 1\\ x^2 & 0 & x^2+y^2 & x^2+z^2 & 1\\ y^2 & x^2+y^2 & 0 & y^2+z^2 & 1\\ z^2 & x^2+z^2 & y^2+z^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|\le \left|\begin{array}{ccccc} \,0 & a^2 & b^2 & c^2 & 1\\ a^2 & 0 & a^2+b^2 & a^2+c^2 & 1\\ b^2 & a^2+b^2 & 0 & b^2+c^2 & 1\\ c^2 & a^2+c^2 & b^2+c^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|\right)$
- An Inequality with Determinants IV $\left(\displaystyle\left|\begin{array}{cccc}\,a&-b&0&0\\0&b&-c&0\\0&0&c&-d\\1&1&1&1+d\end{array}\right|\ge 3\sqrt[3]{4}(abcd)^{5/6}\right)$
- An Inequality with Determinants V $\left(\displaystyle \Delta =\left|\begin{array}{cccc} \,s & \frac{a^2b}{a^3+b} & \frac{b^2c}{b^3+c} & \frac{c^2a}{c^3+a}\\ \frac{a^2b}{a^3+b} & s & \frac{c^2a}{c^3+a} & \frac{b^2c}{b^3+c}\\ \frac{b^2c}{b^3+c} & \frac{c^2a}{c^3+a} & s & \frac{a^2b}{a^3+b}\\ \frac{c^2a}{c^3+a} & \frac{b^2c}{b^3+c} & \frac{a^2b}{a^3+b} & s\end{array}\right|\ge 0\right)$
- An Inequality with Determinants VI $\left(\displaystyle\frac{\Delta_1-\Delta_2}{(b-a)(a-c)(b-c)}\ge 12\sqrt[6]{(abc)^5}\right)$
- An Inequality with Determinants VII $\left(\displaystyle \left|\begin{array}{ccccc} \,0 & a^2 & b^2 & c^2 & 1\\ a^2 & 0 & a^2+b^2 & a^2+c^2 & 1\\ b^2 & a^2+b^2 & 0 & b^2+c^2 & 1\\ c^2 & a^2+c^2 & b^2+c^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right| \le\frac{1}{8}\prod_{cycl}(a+b)^2\right)$
- An Inequality in Reciprocals $\left(\displaystyle\begin{align}\frac{4}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&+4\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\\ &\ge 5\left(\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\right)+\frac{1}{a+b+c}. \end{align}\right)$
- An Inequality in Reciprocals II $\left(\displaystyle \frac{1}{4a+1} + \frac{1}{4b+1} + \frac{6}{2a+2b+1} \ge \frac{4}{3a+b+1} + \frac{4}{3b+a+1}\right)$
- An Inequality in Reciprocals III $\left(\displaystyle \frac{1}{3}\sum_{cycl}\frac{1}{a+b} + \sum_{cycl}\frac{1}{4a+b+c}\ge\sum_{cycl}\left(\frac{1}{a+2b+3c}+\frac{1}{a+3b+2c}\right)\right)$
- A Problem from the Danubius Contest 2016 $\left(\displaystyle\{E(a,b)|\;a\gt 0, b\gt 0\}\subset \left(\frac{11}{10},\frac{10}{9}\right]\right)$
- A Problem from the Danubius-XI Contest $\left(\displaystyle\frac{a+b}{(1+c)^2}+\frac{b+c}{(1+a)^2}+\frac{c+a}{(1+b)^2}\ge\frac{9S}{(9+S)^2}\right)$
- An Inequality with Integrals and Rearrangement $\left(\displaystyle \sum_{cycl}b^2c^3\int_0^a(\cot x)(\arctan x)dx\lt abc(a^3+b^3+c^3)\right)$
- An Inequality with Cot, Cos, and Sin $\left(\displaystyle\sum_{cycl}\cot^2A\,\cot^2B\ge\frac{\displaystyle\sum_{cycl}\cos^2A}{\displaystyle\sum_{cycl}\sin^2A}\right)$
- A Trigonometric Inequality from the RMM $\left(\displaystyle\sin (x+y)\le\sin x\left(\frac{\sin y}{y}\right)^3+\sin y\left(\frac{\sin x}{x}\right)^3\right)$
- An Inequality with Finite Sums $\left(\displaystyle\sum_{k=1}^n\frac{(2+x)^k}{2+x\cdot k}\ge 2^n-1\right)$
- Hung Viet's Inequality $\left(\displaystyle 1+\sum_{cycl}\frac{ab}{a^2-ab+b^2}\ge\frac{3}{2}\cdot\frac{(a+b)(b+c)(c+a)}{a^3+b^3+c^3}\right)$
- Hung Viet's Inequality II $\left(\displaystyle\min_{all}(a-b)^2\le\frac{a^2+b^2+c^2+d^2}{5}\right)$
- Hung Viet's Inequality III $\left(\displaystyle \left(\sum_{cycl}a^4\right)\left(\sum_{cycl}ab^3\right)\ge\left(\sum_{cycl}a^3b\right)\left(\sum_{cycl}a^2b^2\right)\right)$
- Inequality by Calculus $\left(\displaystyle\frac{b^{n+1}-a^{n+1}}{n+1}+\frac{ab(b^{n-1}-a^{n-1})}{n-1}\leq (b-a)\sqrt{2(a^{2n}+b^{2n})}\right)$
- Dorin Marghidanu's Calculus Lemma
$\left(\displaystyle u^v\gt\frac{u}{u+v}\right)$
- An Inequality with Exponents from a Calculus Lemma $\left(\displaystyle a^b\cdot (a+b)^c\cdot (a+b+c)^d\gt\frac{a}{a+b+c+d}\right)$
- An Inequality with Exponents from a Calculus Lemma II $\left(2^a(b+c)^{1-a}+2^b(c+a)^{1-b}+2^c(a+b)^{1-c}\lt 4(a+b+c)\right)$
- An Area Inequality $\left(\displaystyle a^{2} + b^{2} + u^{2} + v^{2} + au + bv \ge \frac{3}{2}\right)$
- A 4-variable Inequality from the RMM $\left(\displaystyle |(a-b)(b-c)(c-a)|\le\sum_{cycl}|a-b||a+c+\eta||b+c+\eta|\right)$
- An Inequality from RMM with Powers of 2 $\left(2^x+2^y+2^z+2^{x+y+z}\gt\sqrt[x+y]{16^{xy}}+\sqrt[y+z]{16^{yz}}+\sqrt[z+x]{16^{zx}}+1\right)$
- A Cycling Inequality with Integrals $\bigg(2bc\Omega(a)+2ca\Omega(b)+2ab\Omega(c)\lt a^2+b^2+c^2$ where $\Omega(t)=\displaystyle\int_{0}^{1}\frac{1-x^2}{1+tx^2+x^4}dx\bigg)$
- A Cycling Inequality with Integrals II $\bigg(\Omega(a,b,c)+\Omega(b,c,a)+\Omega(c,a,b)\le 1,$ where $\Omega(p,q,r)=\displaystyle\int_{0}^{1}\frac{x^p}{1+x^p+x^{q+r}}dx\bigg)$
- An Inequality with Absolute Values $\left(\displaystyle\sum_{cycl}\frac{|a|+|b|}{1-c^2} \ge \sum_{cycl}\frac{2|a|}{1-bc}\right)$
- An Inequality from RMM with a Generic 5 $\left(\displaystyle\sqrt[3]{(2a+5)(2b+5)(2c+5)}\ge\frac{6abc}{ab+bc+ca}+5\right)$
- An Elementary Inequality by Non-elementary Means $\left(12(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3\sin 2a\right)$
- Inequality in Quadrilateral $\left(\displaystyle\sum_{cycl}\sqrt{a^2+b^2+c^2}\gt 2\sqrt{3\cdot AC\cdot BD}\right)$
- Marian Dinca's Refinement of Nesbitt's Inequality $\left(\displaystyle\begin{align}\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}\\&\ge\frac{3\sqrt{3(a^2+b^2+c^2)}}{2(a+b+c)}\\&\ge\frac{3}{2}\end{align}\right)$
- An Inequality in Cyclic Quadrilateral $\left(\displaystyle\frac{R}{8\sqrt{2}}\ge\frac{\sqrt[4]{(abcd)^3}}{(a+b+c+d)^2}\right)$
- An Inequality in Cyclic Quadrilateral II $\left(\displaystyle\sin A\sin B\le (\frac{s}{a}-1)(\frac{s}{b}-1)(\frac{s}{c}-1)(\frac{s}{d}-1)\right)$
- An Inequality in Cyclic Quadrilateral III $\left(\displaystyle\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\cos\frac{D}{2}\le\frac{S^2}{4abcd}\right)$
- An Inequality in Cyclic Quadrilateral IV $\left(a^2-b^2-c^2+d^2+4S\le 2\sqrt{2}(ad+bc)\right)$
- Inequality with Three Linear Constraints $\left(\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 1\right)$
- Inequality with Three Numbers, Not All Zero $\left(\displaystyle\frac{a^2-ab+b^2}{b^2+bc+c^2}+\frac{b^2-bc+c^2}{c^2+ca+a^2}+\frac{c^2-ca+a^2}{a^2+ab+b^2}\ge 1\right)$
- An Easy Inequality with Three Integrals $\left(\displaystyle a^2\int_{0}^{b}\frac{\arctan x}{x}dx + b^2\int_{0}^{c}\frac{\arctan x}{x}dx + c^2\int_{0}^{a}\frac{\arctan x}{x}dx \lt a^3+b^3+c^3\right)$
- Divide And Conquer in Cyclic Sums $\left(\displaystyle \sum_{cycl}c\left(\frac{4a}{b^2}+\frac{3b}{a^2}\right)\ge 12+3\sum_{cycl}\frac{a}{b}\right)$
- Wu's Inequality $\left((x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\ge (xy+yz+zx)^3\right)$
- Dorin Marghidanu's Inequality in Complex Plane $\left(\displaystyle \sum_{cycl}(|(2-n)\cdot z_1+z_2+\ldots+z_n|\ge\sum_{k=1}^n|z_k|\right)$
- Dorin Marghidanu's Inequality in Integer Variables $\left(\displaystyle \frac{m}{\sqrt[m]{1+n}}+\frac{n}{\sqrt[n]{1+m}}\gt\frac{m+n}{2}\right)$
- Dorin Marghidanu's Inequality in Many Variables $\left(\displaystyle\prod_{k=1}^{n}\sqrt[n]{\frac{\displaystyle \sum_{i=1,i\ne k}^{k}a_i}{a_k)}} \ge n-1\right)$
- Dorin Marghidanu's Inequality in Many Variables Plus Two More $\left(\displaystyle G_n\left[p+\frac{r}{a_1},p+\frac{r}{a_2},\ldots,p+\frac{r}{a_n}\right]\ge p+\frac{r}{A_n[a_1,a_2,\ldots,a_n]}\right)$
- Dorin Marghidanu's Inequality with Radicals $\left(\displaystyle\sum_{k=1}^{n}\sqrt[i_k]{x_k} \gt \sqrt[\small{\displaystyle \sum_{k=1}^{n}i_k}]{\prod_{k=1}^{n}x_k}\right)$
- Dorin Marghidanu's Light Elegance in Four Variables $\left(\displaystyle \sum_{cycl}(-a+b+c+d)^2\ge 2(a+b+c+d)-1\right)$
- Dorin Marghidanu's Spanish Problem $\left(\displaystyle n^*\le (n_*)^2\right)$
- Two-Sided Inequality - One Provenance $\left(\displaystyle\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k}+\sqrt{2k+1}}\lt n\lt\sum_{k=1}^{2n(n+1)}\frac{1}{\sqrt{2k-1}+\sqrt{2k}}\right)$
- An Inequality with Factorial $\left(a_1\cdot a_2\cdot\ldots\cdot a_n+(1-a_1)\cdot (2-a_2)\cdot\ldots\cdot (n-a_n)\le n!\right)$
- Wonderful Inequality on Unit Circle $\left(\displaystyle\left(\frac{a+b}{1+ab}\right)^2+\left(\frac{a-b}{1-ab}\right)^2\ge 1\right)$
- Quadratic Function for Solving Inequalities $\left((a^2+3x^2)(b^2+3y^2)(c^2+3z^2)\ge 4(ayz+bzx+cxy+xyz)^2\right)$
- An Inequality Where One Term Is More Equal Than Others $\left(\displaystyle\left(\sum_{k=1}^na_k\right)\left(\sum_{k=1}^n\frac{1}{a_k}\right)\ge n^2+(n-2)^2\right)$
- Complicated Constraint - Simple Inequality $\left(3(a+b)(b+c)(c+a)\ge\frac{\displaystyle 8}{\displaystyle\sqrt[8]{a^3+b^3+c^3}}\right)$
- The power of substitution II: proving an inequality with three variables $\left(\displaystyle\frac{ab}{(a+b)^2}+\frac{bc}{(b+c)^2}+\frac{ca}{(c+a)^2}\le\frac{1}{4}+\frac{4abc}{(a+b)(b+c)(c+a)}\right)$
- Algebraic-Geometric Inequality $\left(\sqrt{x^2-\sqrt{3}xy+y^2} + \sqrt{y^2-\sqrt{2}yz+z^2} \ge \sqrt{z^2-zx+x^2}\right)$
- One Inequality - Two Domains $\left(\displaystyle 3\prod_{cycl}(a^2+ab+b^2)\ge\left(\sum_{cycl}a\right)^2\cdot\left(\sum_{cycl}ab\right)^2\right)$
- Radicals, Radicals, And More Radicals in an Inequality $\bigg(\displaystyle\gamma=\frac{\sqrt[4]{xz}}{\sqrt{x}+\sqrt{z}}.\,$ Prove that $\sqrt{x}+\sqrt{y}+\sqrt{z}\ge 2\gamma\sqrt{3(x+y+z)}\bigg)$
- An Inequality in Triangle and In General $\left(\displaystyle\sum_{cycl}\frac{\cot A\,\cot^3B}{\cot^2B+2\cot^2A}+2\sum_{cycl}\frac{\cot^2A\cot B}{\cot A+2\cot B}\ge 1\right)$
- Dan Sitaru's Cyclic Inequality In Many Variables $\left(\displaystyle a+b+c+d\le \frac{a^5+b^5+c^5+d^5}{abcd}\right)$
- An Inequality on Circumscribed Quadrilateral $\left(s\ge 4R\right)$
- An Inequality with Fractions $\left(\displaystyle m\le\frac{a_1+a_2+\ldots+a_n}{b_1+b_2+\ldots+b_n}\le M\right)$
- An Inequality with Complex Numbers of Unit Length $\left(|a-b|+|a-c|\ge |a+b|+|a+c|\right)$
- An Inequality with Complex Numbers of Unit Length II $\left(|a^2+bc|\ge |b+c|\right)$
- Le Khanh Sy's Problem $\left(xa^2+yb^2+zc^2\ge 2m\right)$
- An Inequality Not in Triangle $\left(\displaystyle\sqrt{a^2+b^2-ab\sqrt{2}}+\sqrt{b^2+c^2-bc\sqrt{3}}+\sqrt{c^2+d^2-\frac{cd(\sqrt{6}+\sqrt{2})}{2}}\ge\sqrt{a^2+d^2}\right)$
- An Acyclic Inequality in Three Variables $\left(\displaystyle \frac{(a^2-bc)^2+(b^2-ca)^2+(c^2-ab)^2}{a^2+b^2+c^2+ab+bc+ca}\geq 3(a-b)(b-c)\right)$
- An Inequality with Areas, Norms, and Complex Numbers $\left(\displaystyle \frac{(ad-bc)(3(a^2+b^2)(c^2+d^2)-4(ad-bc)^2)}{\left((a^2+b^2)(c^2+d^2)\right)^{\frac{3}{2}}}\le 1\right)$
- Darij Grinberg's Inequality In Three Variables $\left(a^2+b^2+c^2+2abc+1\ge 2(ab+bc+ca)\right)$
- Small Change Makes Big Difference $\left(\displaystyle\frac{1}{\displaystyle \sqrt{1+a^2-\frac{(a-b)^2}{2}}}+\frac{1}{\displaystyle \sqrt{1+b^2-\frac{(a-b)^2}{2}}}\ge\frac{2}{\sqrt{1+ab}}\right)$
- Inequality with Two Variables? Think Again $\left(\displaystyle\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le\frac{2}{\sqrt{1+ab}}\right)$
- A Problem From a Mongolian Olympiad for Grade 11 $\left(\displaystyle \frac{a}{3a+2b^3}+ \frac{b}{3b+2c^3}+ \frac{c}{3c+2a^3}\le\frac{1}{5}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\right)$
- Sitaru--Schweitzer Inequality
$\left(\displaystyle \left(\int_{a}^{b}f(x)dx\right)\left(\int_{a}^{b}\frac{1}{f(x)}dx\right)\le\frac{(m+M)^2}{4mM}(b-a)^2\right)$
- Pólya-Szegö Inequality $\left(\displaystyle \frac{\displaystyle \left(\sum_{k=1}^{n}a^2_{k}\right)\left(\sum_{k=1}^{n}b^2_{k}\right)}{\displaystyle\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^2}\le\left(\frac{\displaystyle \sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}}{2}\right)^2\right)$
- Kantorovich Inequality $\left(\displaystyle \left(\sum_{k=1}^{n}\gamma_ku_k^2\right)\left(\sum_{k=1}^{n}\frac{1}{\gamma_k}u_k^2\right)\le\frac{1}{4}\left(\sqrt{\frac{M}{m}}+\sqrt{\frac{m}{M}}\right)^2\left(\sum_{k=1}^{n}u^2_{k}\right)^2\right)$
- Greub-Rheinboldt Inequality $\left(\displaystyle \left(\sum_{k=1}^{n}a_k^2u_k^2\right)\left(\sum_{k=1}^{n}b_k^2u_k^2\right)\le\frac{(M_1M_2+m_1m_2)^2}{4m_1m_2M_1M_2}\left(\sum_{k=1}^{n}a_kb_ku^2_{k}\right)^2\right)$
- An Inequality with Cyclic Sums And Products $\left(\small{\displaystyle \sum_{cycl}\frac{a^2}{(b+c+d+e)(a-b)(a-c)(a-d)(a-e)}\lt\frac{(a+b+c+d+e)^2}{1024abcde}}\right)$
- Problem 1 From the 2016 Pan-African Math Olympiad $\left(\displaystyle \sum_{cycl}\frac{1}{(x+1)^2+y^2+1}\le\frac{1}{2}\right)$
- An Inequality with Integrals and Radicals $\left(\displaystyle \Bigr(\int_0^1 \sqrt[3]{f(x)}dx\Bigr)\Bigr(\int_0^1 \sqrt[5]{f(x)}dx\Bigr)\Bigr(\int_0^1 \sqrt[7]{f(x)}dx\Bigr)\leq 1\right)$
- Twin Inequalities in Four Variables: Twin 1 $\left(\displaystyle (ac+bd)^2\le\left(b\sqrt[5]{ab^4}+d\sqrt[5]{cd^4}\right)\left(a\sqrt[5]{a^4b}+c\sqrt[5]{c^4d}\right)\right)$
- Twin Inequalities in Four Variables: Twin 2 $\left(\displaystyle (a\sqrt[3]{a^2b}+c\sqrt[3]{c^2d})(b\sqrt[3]{ab^2}+d\sqrt[3]{cd^2})\le (a^2+c^2)(b^2+d^2)\right)$
- Simple Inequality with a Variety of Solutions $\left(\displaystyle \sum_{cycl}\left(\frac{\ln x}{\ln y\ln z}+\frac{\ln y}{\ln z\ln x}\right)\ge\frac{18}{\ln (xyz)}\right)$
- A Partly Cyclic Inequality in Four Variables $\left(\displaystyle \sum_{cycl}xe^x\ge (x+y+2)e^{x+y+2}+(z+t-2)\sqrt[3]{e^{z+t-2}}\right)$
- Dan Sitaru's Inequality by Induction $\left(\displaystyle\begin{align}&\small{\frac{3}{a+1}+\frac{3}{b+1}+\frac{2}{c+1}+\frac{1}{d+1}}\\ &\small{\qquad\le 6+\frac{1}{a+b+1}+\frac{1}{a+b+c+1}+\frac{1}{a+b+c+d+1}}\end{align}\right)$
- An Inequality in Three (Or Is It Two) Variables $\left(\displaystyle \frac{(x+y)^2}{(x\sin^2z+y\cos^2z)(x\cos^2z+y\sin^2z)}+\frac{x}{y}+\frac{y}{x}\geq 6\right)$
- An Inequality in Four Weighted Variables $\left(\displaystyle (a+c)^c(b+d)^d(c+d)^{c+d}\le c^cd^d(a+b+c+d)^{c+d}\right)$
- An Inequality in Fractions with Absolute Values $\left(\displaystyle \omega\lt\frac{1}{3}\left(\sum_{cycl}\frac{a|a|-b|b|}{a-b}\right)\lt 2\Omega\right)$
- Inequalities with Double And Triple Integrals $\left(\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\cos\left(\frac{x+y}{2}\right)dxdy\ge\frac{\pi}{2}\right)$
- An Old Inequality $\left(\displaystyle \frac{4ab}{(a+b)^2}\ge\cos\left(\frac{\pi}{2}\cdot\frac{a-b}{a+b}\right)\right)$
- Dan Sitaru's Amazing, Never Ending Inequality $\left(\displaystyle \small{\sum_{cycl}\left(\frac{a}{b}\right)^2\cdot\sum_{cycl}\left(\frac{a}{b}\right)^4\cdot\sum_{cycl}\left(\frac{a}{b}\right)^8\ge\sum_{cycl}\left(\frac{a}{c}\right)\cdot\sum_{cycl}\left(\frac{b}{a}\right)\cdot\sum_{cycl}\left(\frac{b}{c}\right)}\right)$
- Leo Giugiuc's Exercise $\left(\displaystyle x\sin x+x^2\cos x\le 2\sin^2x\right)$
- Another Inequality with Logarithms, But Not Really $\left(\displaystyle \sum_{cycl}\frac{\log_y^3x+\log_z^3y}{\log_y^2x+\log_zx+\log_z^2y}\geq 2\right)$
- An Inequality Solved by Changing Appearances $\left(\displaystyle \sum_{cycl}a^2\cdot\sum_{cycl}\frac{1}{x^2}+\frac{\displaystyle 2\sum_{cycl}ab\cdot\sum_{cycl}x}{xyz}\ge 0\right)$
- Distances to Three Points on a Circle $\left(3\le |z-a|+|z-b|+|z-c|\le 4\right)$
- An Inequality with Powers And Logarithm $\left(\displaystyle \frac{a}{b}+\frac{a^2}{b^2}+\frac{a^3}{b^3}+12\ln b\ge\frac{b}{a}+\frac{b^2}{a^2}+\frac{b^3}{a^3}+12\ln a\right)$
- Four Integrals in One Inequality $\left(\displaystyle \small{\left(\int_a^bxf(x)dx\right)\left(\int_a^bf^2(x)dx\right)\left(\int_a^bx^3f(x)dx\right)\ge\frac{a^2b^2}{b-a}\left(\int_a^bf(x)dx\right)^4}\right)$
- Same Integral, Three Intervals $\left(\displaystyle\small{I(u,v)=\int_u^v\left(\arctan\left(\frac{u\sin x}{v+u\cos x}\right)+\arctan\left(\frac{v\sin x}{u+v\cos x}\right)\right)dx}\right)$
- Dorin Marghidanu's Inequality with Generalization $\left(\displaystyle (x+y)^2+(y+z)^2+(z+x)^2+12ab\le 4(a+b)(x+y+z)\right)$
- Dan Sitaru's Inequality with Three Related Integrals and Derivatives $\left(\displaystyle\small{\left(\int_0^af(x)dx\right)^4\leq \frac{a^8}{60}\left(\int_0^a \left(f'(x)\right)^2 dx\right)\left(\int_0^a \left(f''(x)\right)^2dx\right)}\right)$
- An Inequality in Two Or More Variables $\left(\displaystyle \frac{a}{1+a}+\frac{b}{(1+a)(1+b)}+\frac{c}{(1+a)(1+b)(1+c)}\geq \frac{7}{8}\right)$
- An Inequality in Two Or More Variables II $\left(\displaystyle (a+1)^{a+1}\cdot (b+1)^{b+1}\cdot (c+1)^{c+1}\le e^{a+b+c}\cdot\sqrt{e^{a^2+b^2+c^2}}\right)$
- A Not Quite Cyclic Inequality $\left(\displaystyle \frac{a^2+b^2+c^2}{a+b+c} \le \frac{ab+bc+ca}{a+b+c} + |a-b|+|b-c|\right)$
- Dan Sitaru's Inequality: From Three Variables to Many in Two Ways $\left(\displaystyle a+b+c\ge\frac{3}{2}\right)$
- An Inequality with Sines But Not in a Triangle $\left(\displaystyle \prod_{cycl}\Bigr(a^2\sin \frac{2\pi}{a}+(a+1)^2\sin \frac{2\pi}{a+1}\Bigr)\gt 2^{16}\right)$
- An Inequality with Angles and Integers $\left(\displaystyle k^2\tan \alpha +l^2\tan \beta \geq \frac{2kl}{\sin (\alpha+\beta)}-(k^2+l^2)\cot (\alpha+\beta)\right)$
- Sladjan Stankovik's Inequality In Four Variables $\left(\displaystyle 2\sum_{cycl}a^2-3\frac{\displaystyle \sum_{cycl}a^3}{\displaystyle \sum_{cycl}a}\le\sum_{all}ab\right)$
- An Inequality with Two Pairs of Triplets $\left(\displaystyle (a^2+b^2+c^2)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2(ab+bc+ca)(x+y+z)}{xyz}\ge 0\right)$
- A Refinement of Turkevich's Inequality $\left(\displaystyle a^2+b^2+c^2+d^2+\frac{32abcd}{(a+b+c+d)^2}\ge \sum_{sym}ab\right)$
- Dan Sitaru's Exercise with Pi and Ln $\left(\displaystyle \small{\arctan(z-x)+\arctan(z-y)+\arctan(y-x) \lt \frac{\pi}{2} - \ln(2)}\right)$
- Leo Giugiuc's Cyclic Quickie in Four Variables $\left(3(xyz+yzt+ztx+txy)^2\ge 8(xy+xz+xt+yz+yz+zt)xyzt\right)$
- Dan Sitaru's Cyclic Inequality in Four Variables $\left(\displaystyle \sum_{cycl}\frac{a^7}{bcd+a^3}\ge 2abcd\right)$
- A Not Quite Cyclic Inequality from Tibet $\left((x+y)^2(z^2+xz+x^2+xy+y^2+yz)^2\ge 8(xy+yz+zx)^2(x^2+y^2)\right)$
- An inequality in 2+2 variables from SSMA magazine $\left(\displaystyle k^2\tan\alpha+l^2\tan\beta\ge\frac{2kl}{\sin (\alpha+\beta)}-(k^2+l^2)\cot(\alpha+\beta)\right)$
- Kunihiko Chikaya's Inequality with Parameter $\bigg(p \ge 2.$ Prove $\displaystyle \sum_{cycl}\frac{a}{\sqrt{ap+b}} \le \sqrt{\frac{3(a+b+c)}{p+1}}\bigg)$
- Dorin Marghidanu's Permuted Inequality $\left(\displaystyle \sum_{k=1}^n\left(a_k+\frac{1}{a_{\sigma(k)}}\right)^p\ge \frac{(s^2+n^2)^p}{n^{p-1}s^p}\right)$
- An Inequality Involving Arithmetic And Geometric Means $\left(\displaystyle\sum_{cycl}\frac{1}{a^4+b^4+c^4+abcd}\le \frac{1}{abcd}\right)$
- Dorin Marghidanu's Sums and Products $\left(\displaystyle \sum_{k=1}^n\frac{a_k}{P_kS_k}\ge\frac{n^n}{\displaystyle (n-1)S^{n-1}}\right)$
- Simple Nameless Inequality $\left(\displaystyle \sum_{k=1}^n\frac{S}{S_k}\ge\frac{n^2}{n-1}\right)$
- Volume Inequality in Tetrahedron $\left(OA\cdot OB\cdot OC\ge 27xyz\right)$
- Inequality in Convex Quadrilateral $\left(\displaystyle\frac{\displaystyle \sum_{cycl}\sqrt{b+c+d-a}}{a+b+c+d}\ge\sqrt{\frac{2(a+b+c+d)}{a^2+b^2+c^2+d^2}}\right)$
- Dan Sitaru's Inequality with a Double Integral $\left(\displaystyle\begin{align}&\small{\int_0^1\int_0^1\sqrt{\left(m^2\sqrt{mnf(x)f(y)}+f^2(x)\right)\left(n^2\sqrt{mnf(x)f(y)}+f^2(y)\right)}dxdy}\\ &\qquad\qquad\qquad\qquad\small{\ge (m+n)\int_0^1f(x)dx}.\end{align}\right)$
- Cute Exercise by Dorin Marghidanu $\left(\displaystyle \sum_{k=1}^n\frac{2k-1}{\sqrt[2k-1]{a_k}}\ge\frac{n^2}{\sqrt[n^2]{a_1a_2\ldots a_n}}\right)$
- A Little of Algebra for an Inequality, A Little of Calculus for a Generalization $\left(\displaystyle \frac{a^{n+1}-b^{n+1}}{a-b}\cdot\frac{b^{n+1}-c^{n+1}}{b-c}\cdot\frac{c^{n+1}-a^{n+1}}{c-a}\gt (n+1)^3(abc)^n\right)$
- An Inequality with Central Binomials $\left(\displaystyle \sqrt{2}\le\sqrt[n(n+1)]{{2\choose 1}{4\choose 2}\cdots{2k\choose k}\cdots{2n\choose n}}\lt 2\right)$
- A Simple Inequality with Many Variables $\left(\displaystyle \sum_{k=1}^n\sqrt{\frac{a_k+a_{k+1}}{a_{k+2}}}\ge n\sqrt{2}\right)$
- Cyclic Inequality in Four Variables $\left(\displaystyle \frac{3}{4}\sum_{cycl}\frac{a^3}{bcd}\ge 1+\frac{\displaystyle 3\sum_{cycl}a^2}{\displaystyle \sum_{all}ab}\right)$
- Cyclic Inequality in Four Variables By D. Sitaru $\left(\displaystyle \sum_{cycl}\frac{a^7}{a^3+bcd}\ge 2abcd\right)$
- Lorian Saceanu's Inequality with Many Variables $\left(\displaystyle \frac{1}{2}\left(\sqrt{ab}+\frac{1}{\sqrt{ab}}\right)\left(\sum_{i=1}^na_kb_k\right)\ge\sqrt{\left(\sum_{i=1}^na_k^2\right)\left(\sum_{i=1}^nb_k^2\right)}\right)$
- A True Algebraic-Geometric Inequality $\left(\displaystyle \small{\sqrt{\sum_{k=1}^n(2a_k-b_k)^2}+\sqrt{\sum_{k=1}^n(2b_k-a_k)^2}\ge\sqrt{\sum_{k=1}^na_k^2}+\sqrt{\sum_{k=1}^nb_k^2}}\right).$
- Leo Giugiuc's Cyclic Inequality in Square Roots $\left(\displaystyle \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\right)$
Working backwards
Substitution
Equivalence classes
Invariants
Transformation
Angle chasing
The method of angle chasing is explained on a separate page.
Counterexample
Counterexample is an example with a negative connotation. Whereas an example may be used to support or illustrate a claim, a counterexample is used to refute an assertion. How an example is being used often depends on the purpose or the formulation. For example, the word "nth" is an example of an English word without a vowel. It is a counterexample to the claim that every English word contains a vowel.
For a few more words and many additional examples see a separate file.
Coordinates and Analytic Methods
- The Three Jugs Problem
- Cabart's Collinearity
- Complex Numbers and Geometry
- Doodling and Miracles
- Two Butterflies Theorem
- Carnot's Theorem
- Semicircle on Square
- Eyeball Theorem Rectified
- RomanoNorwegian Inequality
- Another Equilateral Triangle in Napoleon's Configuration
- Circles in Barycentric Coordinates
- Barycenter of Cevian Triangle
Infinite Descent
Fixed point
Projective and affine methods
- Gergonne in Ellipse
- Expressions Invariant under Projective Mappings
- Cevians and Conic through Six Points on Sidelines of Triangle
- A Matter of Collinearity: Metamorphosis of a Problem
- A Property of Perspective Triangles
- Two Circles, Tangents and Two Collinearities
- Projective Invariants For Pascal And Desargues
Shortcuts
Very often, especially when a problem has been posed at an olympiad or in math circles, a solution that comes to mind first is not necessarily the best - the easiest to follows through. At least occasionaly, there are shortcuts that simplify solving the problem considerably.
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