# Hung Nguyen Viet's Inequality with a Constraint

### Solution

First we homogenize the equation, making it equivalent to

$\displaystyle \left(\sum_{cycl}x^2\right)\left[\left(\sum_{cycl}x^2\right)^2+2\left(\sum_{cycl}xy\right)^2\right]\ge \left(\sum_{cycl}x^3+6xyz\right)^2.$

As a strategy, let, WLOG, $x+y+z=3,$ then $xy+yz+zx=3(1-t^2),$ $0\le t\le 1,$ and $\max(xyz)=2t^3-3t^2+1.$ Further,

\begin{align} x^2+y^2+z^2&=(x+y+z)^2-2(xy+yz+zx)\\ &=9-6(1-t^2)=3(2t^2+1) \end{align}

and

\begin{align} \left(\sum_{cycl}x^2\right)^2+2\left(\sum_{cycl}xy\right)^2&=[3(2t^2+1)]^2+2[3(1-t^2)]^2\\ &=2t^4+1. \end{align}

In addition, $x^3+y^3+z^3=3t^2+xyz.$ The problem thus reduces to proving

$(2t^2+1)(2t^4+1)\ge (3t^2+xyz)^2.$

Suffice it to prove that

$(2t^2+1)(2t^4+1)\ge (2t^3+1)^2,$

which is equivalent to, $t^2(1-t)^2\ge 0$ which is obvious.

Returning to the original problem, equality is attained for $x=y=z=\displaystyle \frac{1}{\sqrt{3}}$ or $(x,y,z)=(1,0,0)$ and permutations.

### Acknowledgment

Leo Giugiuc has kindly communicated to me that problem by Hung Nguyen Viet, along with a solution of his. The problem was originally posted on January 18, 2018 at the Olimpiada pe Şcoală (The School Yard Olympiad) facebook group.

Copyright © 1996-2018 Alexander Bogomolny

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