Another Problem from the 2016 Danubius Contest
Problem
Solution 1
The inequality is equivalent to
$\displaystyle 12+4\sum_{cycl}a^2+\sum_{cycl}a^2b^2\le 8+4\sum_{cycl}a^2+2\sum_{cycl}a^2b^2+a^2b^2c^2,$
or,
$\displaystyle 4\le a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2.$
let's denote $bc=x,\,$ $ca=y,\,$ and $ab=z.\,$ Then $x,y,z\gt 0,\,$ $x+y+z=3\,$ and $xyz=a^2b^2c^2.\,$ We have to show that
$x^2+y^2+z^2+xyz\ge 4.$
We can homogenize the inequality:
$\displaystyle 4\left(\frac{x+y+z}{3}\right)^3\le (x^2+y^2+z^2)\left(\frac{x+y+z}{3}\right)+xyz,$
reducing it to $5S_3-3s+3xyz,\,$ where $s_3=x^3+y^3+z^3\,$ and $s=\displaystyle \sum_{cycl}xy(x+y).\,$ From Schur's inequality, $S_3-s+3xyz\ge 0\,$ and, from the well-known inequality $u^3+v^3\ge uv(u+v),\,$ $4S_3-2s\ge 0.\,$ Adding the two up gives the required inequality.
Solution 2
$\displaystyle\sum_{cycl}\frac{1}{a^2+2}\le 1\,$ is equivalent to $\displaystyle\sum_{cycl}\frac{2}{a^2+2}\le 2\,$ and, further, to $\displaystyle\sum_{cycl}\left(1-\frac{a^2}{a^2+2}\right)\le 2\,$ which is $\displaystyle\sum_{cycl}\frac{a^2}{a^2+2}\ge 1.\,$ To prove this we can use Bergstrom's inequality:
$\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{a^2+2}&\ge\frac{(a+b+c)^2}{a^2+b^2+c^2+6}\\ &=\frac{(a^2+b^2+c^2)+2(ab+bc+ca)}{a^2+b^2+c^2+6}\\ &=\frac{(a^2+b^2+c^2)+2\cdot 3}{a^2+b^2+c^2+6}=1. \end{align}$Solution 3
let's denote $bc=x,\,$ $ca=y,\,$ and $ab=z.\,$ Then $x,y,z\gt 0,\,$ $x+y+z=3.\,$ We have to show that
$x^2+y^2+z^2+xyz\ge 4.$
Assuming $x=\min\{x,y,z\},\,$ $x\le 1,\,$ and we have
$\displaystyle \begin{align} x^2+y^2+z^2+xyz-4 &= x^2+(y+z)^2+yz(x-2)-4\\ &\ge x^2+(y+z)^2+\frac{1}{4}(y+z)^2(x-2)-4\\ &\ge x^2+\frac{x+2}{4}(y+z)^2-4\\ &=x^2+{x+2}{4}(3-x)^2-4\\ &=\frac{1}{4}(x-1)^2(x+2)\ge 0. \end{align}$
Equality occurs for $a=b=c=1.$
Solution 4
$\displaystyle c=\frac{3-ab}{a+b}.$
$\displaystyle\begin{align} RHS-LHS &= \frac{c^2+1}{c^2+2}-\frac{a^2+b^2+4}{(a^2+2)(b^2+2)}\\ &=\frac{(3-ab)^2+(a+b)^2}{(3-ab)^2+2(a+b)^2}-\frac{a^2+b^2+4}{(a^2+2)(b^2+2)}. \end{align}$
Denominators being positive, we'll focus on the numerators:
$\begin{align} &(2a^2+2b^2+4)((3-ab)^2+(a+b)^2)\\ &\qquad\qquad\qquad\qquad-(a^2+b^2+4)((3-ab)^2+2(a+b)^2)\\ &\qquad\qquad\qquad\qquad+a^2b^2((3-ab)^2+(a+b)^2)\\ &=(a^2+b^2)(3-ab)^2-4(a^2+b^2)\\ &\qquad\qquad\qquad\qquad-8ab+a^2b^2(a^2+b^2-4ab+9+a^2b^2)\\ &=(a^2+b^2)(a^2b^2-6ab+9-4+a^2b^2)\\ &\qquad\qquad\qquad\qquad-ab(a^3b^3-4a^2b^2+9ab-8)\\ &=(a^2+b^2)(2a^2b^2-6ab+4)+ab(a^3b^3-4a^2b^2+9ab-6)\\ &\qquad\qquad\qquad\qquad+(a-b)^2\\ &=(a^2+b^2)(ab-1)(ab-2)+ab(ab-1)(a^2b^2-3ab+6)\\ &\qquad\qquad\qquad\qquad+(a-b)^2\\ &\ge 2\cdot 2ab(ab-1)(ab-2)+ab(ab-1)(a^2b^2-3ab+6)\\ &\qquad\qquad\qquad\qquad+(a-b)^2\\ &=ab(ab-1)(a^2b^2+ab-2)+(a-b)^2\\ &=ab(ab-1)(ab-1)(ab+2)+(a-b)^2\\ &=ab(ab-1)^2(ab+2)+(a-b)^2\\ &\ge 0. \end{align}$
Solution 5
We make the same substitution as in Solutions 1 and 3 to obtain the problem of proving
$xyz+x^2+y^2+z^2-4\ge 0,$
provided $x+y+z=3\,$ and $x,y,z\gt 0.\,$ Let $x=k+m,\,$ $y=k-m,\,$ $z=3-x-y=3-2k.\,$ The positivity of $x,y,\,$ and $z\,$ implies
$\displaystyle \frac{3}{2}\gt k\gt 0\,$ and $k\gt m\gt -k.$
We have
$\begin{align} f(k,m) &:= (k+m)(k-m)(3-2k)+(k+m)^2\\ &\qquad\qquad\qquad\qquad+(k-m)^2+(3-2k)^2-4\\ &=[(5-2k)(k-1)^2]+[(2k-1)m^2]. \end{align}$
Note: The first term (the first square bracket) is non-negative over the alowed range of $k,\,$ becoming $0\,$ at $k=1.\,$ The sign of the second term depends on $k.\,$ Let us consider two cases:
$\mathbf{\text{Case 1: }\displaystyle \frac{1}{2}\gt k\gt 0}.\,$ In this case, the second term is negative and for a fixed value f $k\,$ is a monotonically decreasing function of $m^2.\,$ Moreover, $m^2\lt k^2\,$ implies
$\displaystyle \begin{align} f&\gt (5-2k)(k-1)^2+(2k-1)k^2\\ &=\frac{(4k-3)^2}{2}+\frac{1}{2}\\ &\gt\frac{1}{2}. \end{align}$
$\mathbf{\text{Case 2: }\displaystyle \frac{3}{2}\gt k\ge\frac{1}{2}}.\,$ In this case, the second term is non-negative and takes value $0\,$ when $\displaystyle k=\frac{1}{2}\,$ or $m=0.\,$ Thus, $f\,$ takes minimum value of $0\,$ when $k=1\,$ andf $m=0.$
Solution 6
From $ab+bc+ca=3,\,$ $a^2+b^2+c^2\ge 3,\,$ by the AM-GM inequality. If we use 3D polar coordinates, then the problem reduces to finding the maximum of
$\displaystyle \frac{1}{\rho^2\sin^2\theta\cos^2\varphi}+\frac{1}{\rho^2\sin^2\theta\sin^2\varphi}+\frac{1}{\rho^2\cos^2\theta},$
where, as we just observed $\rho^2\ge 3.\,$ The above function decreases as $rho\,$ grows so that it achieves its maximum on the sphere $rho^2=3\,$ - for the minimal value of $rho.$
Further, from $ab+bc+ca=3\,$ and $a^2+b^2+c^2=3\,$ it follows that $(a+b+c)^2=9,\,$ i.e., $a+b+c=3,\,$ the tangent plane to the sphere $a^2+b^2+c^2\,$ at the point $(1,1,1).\,$ This proves the required inequality as
$\displaystyle \frac{1+a^2}+\frac{1+b^2}+\frac{1+c^2}\le\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1.$
Illustration
Acknowledgment
Leo Giugiuc has kindly posted at the CutTheKnotMath facebook page the above problem which he coauthored with Dan Sitaru. The problem has been included at the 2016 Danubius contest. Solution 1 is by the authors; Solution 2 is by Marian Daniel Vasile; Solution 3 is by Vasile Cîrtoaje; Solution 4 is by Srinivas Vemuri; Solution 5 is by Amit Itagi; Solution 6 is by Andrea Acquaviva. The illustration is by N. N. Taleb.
A few variants of this problem have been discussed elsewhere.
- A Cyclic But Not Symmetric Inequality in Four Variables $\left(\displaystyle 5(a+b+c+d)+\frac{26}{abc+bcd+cda+dab}\ge 26.5\right)$
- An Inequality with Constraint $\left((x+1)(y+1)(z+1)\ge 4xyz\right)$
- An Inequality with Constraints II $\left(\displaystyle abc+\frac{2}{ab+bc+ca}\ge\frac{5}{a^2+b^2+c^2}\right)$
- An Inequality with Constraint III $\left(\displaystyle \frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\ge 3\right)$
- An Inequality with Constraint IV $\left(\displaystyle\sum_{k=1}^{n}\sqrt{x_k}\ge (n-1)\sum_{k=1}^{n}\frac{1}{\sqrt{x_k}}\right)$
- An Inequality with Constraint VII $\left(|(2x+3y-5z)-3(x+y-5z)|=|-x+10z|\le\sqrt{101}\right)$
- An Inequality with Constraint VIII $\left(\sqrt{24a+1}+\sqrt{24b+1}+\sqrt{24c+1}\ge 15\right)$
- An Inequality with Constraint IX $\left(x^2+y^2\ge x+y\right)$
- An Inequality with Constraint X $\left((x+y+p+q)-(x+y)(p+q)\ge 1\right)$
- Problem 11804 from the AMM $\left(10|x^3 + y^3 + z^3 - 1| \le 9|x^5 + y^5 + z^5 - 1|\right)$
- Sladjan Stankovik's Inequality With Constraint $\left(abc+bcd+cda+dab-abcd\le\displaystyle \frac{27}{16}\right)$
- An Inequality with Constraint XII $\left(abcd\ge ab+bc+cd+da+ac+bd-5\right)$
- An Inequality with Constraint XIV $\left(\small{64(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \le 3(a+b+c)^6}\right)$
- An Inequality with Constraint XVII $\left(a^3+b^3+c^3\ge 0\right)$
- An Inequality with Constraint in Four Variables II $\left(a^3+b^3+c^3+d^3 + 6abcd \ge 10\right)$
- An Inequality with Constraint in Four Variables III $\left(\displaystyle\small{abcd+\frac{15}{2(ab+ac+ad+bc+bd+cd)}\ge\frac{9}{a^2+b^2+c^2+d^2}}\right)$
- An Inequality with Constraint in Four Variables V $\left(\displaystyle 5\sum \frac{abc}{\sqrt[3]{(1+a^3)(1+b^3)(1+c^3)}}\leq 4\right)$
- An Inequality with Constraint in Four Variables VI $\left(\displaystyle \sum_{cycl}a^2+6\cdot\frac{\displaystyle \sum_{cycl}abc}{\displaystyle \sum_{cycl}a}\ge\frac{5}{3}\sum_{sym}ab\right)$
- A Cyclic Inequality in Three Variables with Constraint $\left(\displaystyle a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}+2abc=1\right)$
- Dorin Marghidanu's Cyclic Inequality with Constraint $\left(\displaystyle 2a^2-2\sqrt{2}(b+c)a+3b^2+4c^2-2\sqrt{bc}\gt 0\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints $\left(\displaystyle \frac{1}{\sqrt{a+b^2}}+ \frac{1}{\sqrt{b+c^2}}+ \frac{1}{\sqrt{c+a^2}}\ge\frac{1}{\sqrt{a+b+c}}\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints II $\left(\displaystyle \sum_{cycl}\frac{\displaystyle \frac{x}{y}+1+\frac{y}{x}}{\displaystyle \frac{1}{x}+\frac{1}{y}}\le 9\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints III $\left(\displaystyle 12+\sum_{cycl}\left(\sqrt{\frac{x^3}{y}}+\sqrt{\frac{x^3}{y}}\right)\ge 8(x+y+z)\right)$
- Inequality with Constraint from Dan Sitaru's Math Phenomenon $\left(\displaystyle b+2a+20\ge 2\sum_{cycl}\frac{a^2+ab+b^2}{a+b}\ge b+2c+20\right)$
- Another Problem from the 2016 Danubius Contest
- Gireaux's Theorem (If a continuous function of several variables is defined on a hyperbrick and is convex in each of the variables, it attains its maximum at one of the corners)
- An Inequality with a Parameter and a Constraint $\left(\displaystyle a^4+b^4+c^4+\lambda abc\le\frac{\lambda +1}{27}\right)$
- Unsolved Problem from Crux Solved $\left(a_1a_2a_3a_4a_5a_6\le\displaystyle \frac{5}{2}\right)$
- An Inequality With Six Variables and Constraints Find the range of $\left(a^2+b^2+c^2+d^2+e^2+f^2\right)$
- Cubes Constrained $\left(3(a^4+b^4)+2a^4b^4\le 8\right)$
- Dorin Marghidanu's Inequality with Constraint $\left(\displaystyle \frac{1}{a_1+1}+\frac{2}{2a_2+1}+\frac{3}{3a_3+1}\ge 4\right)$
- Dan Sitaru's Integral Inequality with Powers of a Function $\left(\displaystyle\left(\int_0^1f^5(x)dx\right)\left(\int_0^1f^7(x)dx\right)\left(\int_0^1f^9(x)dx\right)\ge 2\right)$
- Michael Rozenberg's Inequality in Three Variables with Constraints $\left(\displaystyle 4\sum_{cycl}ab(a^2+b^2)\ge\sum_{cycl}a^4+5\sum_{cycl}a^2b^2+2abc\sum_{cycl}a\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints IV $\left(\displaystyle \frac{(4x^2y^2+1)(36y^2z^2+1)(9x^2z^2+1)}{2304x^2y^2z^2}\geq \frac{1}{(x+2y+3z)^2}\right)$
- Refinement on Dan Sitaru's Cyclic Inequality In Three Variables $\left(\displaystyle \frac{(4x^2y^2+1)(36y^2z^2+1)(9x^2z^2+1)}{2304x^2y^2z^2}\geq \frac{1}{3\sqrt{3}}\right)$
- An Inequality with Arbitrary Roots $\left(\displaystyle \sum_{cycl}\left(\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\right)\lt 18\right)$
- Leo Giugiuc's Inequality with Constraint $\left(\displaystyle 2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\le ab+bc+ca\right)$
- Problem From the 2016 IMO Shortlist $\left(\displaystyle \sqrt[3]{(a^2+1)(b^2+1)(c^2+1)}\le\left(\frac{a+b+c}{3}\right)^2+1\right)$
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots $\left(\displaystyle \sum_{cycl}\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}\le\frac{4}{5}\right)$
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots II $\left(\displaystyle \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+\sqrt[3]{d}\le\sqrt[3]{abcd}\right)$
- A Simplified Version of Leo Giugiuc's Inequality from the AMM $\left(\displaystyle a^3+b^3+c^3\ge 3\right)$
- Kunihiko Chikaya's Inequality $\displaystyle \small{\left(\frac{(a^{10}-b^{10})(b^{10}-c^{10})(c^{10}-a^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge\frac{125}{3}[(a-b)^3+(b-c)^3+(c-a)^3]\right)}$
- A Cyclic Inequality on [-1,1] $\left(xy+yz+zx\ge 1\right)$
- An Inequality with Two Triples of Variables $\left(\displaystyle\sum_{cycl}ux\ge\sqrt{\left(\sum_{cycl}xy\right)\left(2\sum_{cycl}uv-\sum_{cycl}u^2\right)}\right)$
- 6th European Mathematical Cup (2017), Junior Problem 4 $\left(x^3 - (y^2 + yz + z^2)x + y^2z + yz^2 \le 3\sqrt{3}\right)$
- Dorin Marghidanu's Example $\left(\displaystyle\frac{\displaystyle\frac{1}{b_1}+\frac{2}{b_2}+\frac{3}{b_3}}{1+2+3}\ge\frac{1+2+3}{b_1+2b_2+3b_3}\right)$
- A Trigonometric Inequality with Ordered Triple of Variables $\left((x+y)\sin x+(x-z)\sin y\lt (y+z)\sin x\right)$
- Three Variables, Three Constraints, Two Inequalities (Only One to Prove) - by Leo Giugiuc $\bigg(a+b+c=0$ and $a^2+b^2+c^2\ge 2$ Prove that $abc\ge 0\bigg)$
- Hung Nguyen Viet's Inequality with a Constraint $\left(1+2(xy+yz+zx)^2\ge (x^3+y^3+z^3+6xyz)^2\right)$
- A Cyclic Inequality by Seyran Ibrahimov $\left(\displaystyle \sum_{cycl}\frac{x}{y^4+y^2z^2+z^4}\le\frac{1}{(xyz)^2}\right)$
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints V $\left(\displaystyle \frac{1}{\sqrt{ab(a+b)}}+\frac{1}{\sqrt{bc(b+c)}}+\frac{1}{\sqrt{ca(c+a)}}\le 3+\frac{a+b+c}{abc}\right)$
- Cyclic Inequality In Three Variables From Kvant $\left(\displaystyle \frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}\le 2\right)$
- Cyclic Inequality In Three Variables From Vietnam by Rearrangement $\left(\displaystyle \frac{x^3+y^3}{y^2+z^2}+\frac{y^3+z^3}{z^2+x^2}+\frac{z^3+x^3}{x^2+y^2}\le 3\right)$
- A Few Variants of a Popular Inequality And a Generalization $\left(\displaystyle \frac{1}{(a+b)^2+4}+\frac{1}{(b+c)^2+4}+\frac{1}{(c+a)^2+4}\le \frac{3}{8}\right)$
- Two Constraints, One Inequality by Qing Song $\left(|a|+|b|+|c|\ge 6\right)$
- A Moscow Olympiad Question with Two Inequalities $\left(\displaystyle b^2\gt 4ac\right)$ A Problem form the Short List of the 2018 JBMO $\left(ab^3+bc^3+cd^3+da^3\ge a^2b^2+b^2c^2+c^2d^2+d^2a^2\right)$
- An Inequality from a Mongolian Exam $\left(\displaystyle 2\sum_{i=1}^{2n-1}(x_i-A)^2\ge \sum_{i=1}^{2n-1}(x_i-x_n)^2\right)$
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