An Inequality with Complex Numbers of Unit Length
What Might That Be About?
Problem
Solution 1
The situation is illustrated with the above applet. A worthy observation is that the condition $|a+b+c|\lt 1\,$ is equivalent to claiming that $a,b,c\,$ do not lie in a half-plane whose boundary (a straight line passes through the origin. WLOG, assume that $a=1,\,$ $b=\cos s+i\sin s,\,$ and $c=\cos t-i\sin t.\,$ The aforementioned condition of not belonging to a half-plane is expressed in terms of the following inequalities: $0\lt s,t\lt \pi\,$ and $s+t\gt\pi.$
Now, $\displaystyle |a-b|=2\sin\frac{s}{2},\,$ $\displaystyle |a+b|=2\cos\frac{s}{2},\,$ $\displaystyle |a-c|=2\sin\frac{t}{2},\,$ $\displaystyle |a+c|=2\cos\frac{t}{2}.\,$ We have to prove that
$\displaystyle \sin\frac{s}{2}+\sin\frac{t}{2}\gt \cos\frac{s}{2}+\cos\frac{t}{2}.$
Note that , for $s+t=\pi\,$ the inequality becomes equality.
We proceed with two fundamental identities:
$\displaystyle\begin{align} \sin\alpha+\sin\beta&=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \cos\alpha+\cos\beta&=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}. \end{align}$
Note that, with $0\lt s,t\lt\pi,\,$ we may be certain that $\displaystyle \frac{s-t}{4}\ne\frac{\pi}{2},$ so that $\displaystyle\cos\frac{s-t}{4}\ne 0.\,$ The inequality to prove becomes $\displaystyle\sin\frac{s+t}{4}\gt\cos\frac{s+t}{4},\,$ or $\displaystyle\tan\frac{s+t}{4}\gt 1,\,$ which is true because $\tan (x),$ is increasing on $\displaystyle \left[0,\frac{\pi}{2}\right),\,$ so that $\displaystyle\tan\frac{s+t}{4}\gt \tan\frac{\pi}{4}=1.$
Solution 2
Let zero be represented by $O$. Let the angle that $\overrightarrow{OA}$ needs to be rotated by to align with $\overrightarrow{OB}$ be $\alpha$. The rotation could be in clockwise or anti-clockwise sense. We choose the sense that ensures $0 \leq \alpha \leq \pi$. We define $\beta$ analogously as the angle between $\overrightarrow{OA}$ and $\overrightarrow{OC}$ (with $0 \leq \beta \leq \pi$).
Thus,
$\begin{align}P &\overset{\Delta}{=} |a-b|+|a-c|-|a+b|-|a+c| \\ \nonumber &=2 \left[ \sin \left( \frac{\alpha}{2} \right) + \sin\left(\frac{\beta}{2}\right) - \cos\left(\frac{\alpha}{2}\right) -\cos\left(\frac{\beta}{2}\right) \right] \\ \nonumber &=2\sqrt{2}\left[\sin\left(\frac{\alpha}{2}-\frac{\pi}{4}\right) + \sin\left(\frac{\beta}{2}-\frac{\pi}{4}\right) \right] \\ \nonumber &=4\sqrt{2} \sin\left(\frac{\alpha+\beta}{4}-\frac{\pi}{4}\right) \cos\left(\frac{\alpha-\beta}{4}\right) \end{align}$
We need to prove that $P \geq 0$. The argument of the cosine term is limited to $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$. Thus, the cosine term is non-negative. The argument of the sine term is also limited to $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$. Thus, the sine term is negative only if $(\alpha+\beta) < \pi$ (condition Q).
The constraint can be written as
$\begin{align}&|a+b+c|^2 \leq 1 \\ \nonumber & (1+\cos\alpha+\cos\beta)^2+(\sin\alpha \pm\sin\beta)^2 \leq 1 \\ \nonumber & 3 + 2\cos\alpha + 2\cos\beta + 2\cos (\alpha \mp\beta)] \leq 1 \\ \nonumber & 3 + 4\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)+4\cos^2 \left(\frac{\alpha \mp \beta}{2}\right) -2 \leq 1 \\ \nonumber & \cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right)+\cos^2 \left(\frac{\alpha \mp \beta}{2}\right) \leq 0 \\ \nonumber \end{align}$
The two possibilities $\pm$ arise because the rotational sense used in the definition of $\alpha$ and $\beta$ could be the same or opposite.
Case 1 (Same rotational sense):
$\begin{align}&\cos\left(\frac{\alpha-\beta}{2}\right) \left[\cos\left(\frac{\alpha+\beta}{2}\right) + \cos\left(\frac{\alpha-\beta}{2}\right) \right] \leq 0 \\ \nonumber &\cos\left(\frac{\alpha-\beta}{2}\right) \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \leq 0 \end{align}$
The argument of all the cosine terms lie either in the first or the fourth quadrant where the cosine is non negative. Hence, this condition is satisfied only if either or both $\alpha$ and $\beta$ equal $\pi$. Thus, condition Q is not true and $P \geq 0$ in this case.
Case 2 (Opposite rotational sense):
$\begin{align}&\cos\left(\frac{\alpha+\beta}{2}\right) \left[\cos\left(\frac{\alpha-\beta}{2}\right) + \cos\left(\frac{\alpha+\beta}{2}\right) \right] \leq 0 \\ \nonumber &\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha}{2}\right) \cos\left(\frac{\beta}{2}\right) \leq 0 \end{align}$
The arguments on the second and third cosine terms lie in the first or the fourth quadrant and the condition for either of the terms being equal to zero reduces to case 1. However the first cosine term can be non-positive, even with neither $\alpha$ nor $\beta$ being equal to $\pi$, if the argument is in the second quadrant. This condition is $2\pi \gt \alpha+\beta \geq \pi$. This condition ensures that condition Q fails and $P \geq 0$ in this case.
Solution 3
Consider \( a, b, c \) as elements of \( \mathbb{R}^2 \). Then \( \| a+b+c \| \leq 1 \) means that \( (a+b+c)\cdot a \leq 1 \), and with \( a\cdot a = 1 \) obtain \( (b+c)\cdot a \leq 0 \). Furthermore \( (b+c)\cdot b \geq 0 \) and \( (b+c)\cdot c \geq 0 \), so \( b,c,-a \) lie in the same half-space \( H_{b+c} = \{ x \mid x\cdot (b+c) \geq 0 \} \).
On the other hand, there exists a (filled) ellipse \( E = \{ x \mid \| x - b \| + \| x - c \| \leq R \} \) with focal points \( b, c \) which contains only the part of the unit circle on the half space \( H_{b+c} \). That is, \( S^1 \cap H_{b+c} = S^1 \cap E \).
If \( (b+c)\cdot a = 0 \), then \( a, -a \in \partial E \) and we have equality \( \| a - b \| + \| a - c \| = \| a + b\| + \| a + c\| \). Otherwise \( (b+c)\cdot a < 0 \) and then \( -a \in S^1 \cap H_{b+c} = S^1 \cap E \) and \( a \notin E \). In particular, \( \| a - b \| + \| a - c \| > R \geq \| - a - b \| + \| -a - c \| \).
Illustration
Acknowledgment
The problem which is the invention of Leo Giugiuc and Kadir Altintas, has been posted by Leo Giugiuc on the CutTheKnotMath facebook page. That's an elegantly refreshing problem. Solution 1 is by the two authors; Solution 2 s by Amit Itagi; Solution 3 is by Sophie L. Muller. The illustrated solution is by N. N. Taleb.
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- 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
- 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 $\left(\displaystyle \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\ge 2\sqrt{\frac{(x+y)(y+z)(z+x)}{xy+yz+zx}}\right)$
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