Problem 29: Useful inequality by Vasile Cirtoaje

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by henderson, Jun 29, 2016, 8:43 PM

$\bf\color{red} Problem \ 29$ Let $a,b,c$ be positive real numbers. Prove that
$4(x+y+z)^3\geq 27(x^2y+y^2z+z^2x+xyz).$ $\bf\color{green}Solution \$
Without loss of generality, let's assume that $y$ is the median of $\{ x,y,z\}.$ So, $(y-x)(y-z)\leq 0$ $\Longrightarrow$ $z(y-x)(y-z)\leq 0$ $\Longrightarrow$ $y^2z+z^2x\leq xyz+yz^2.$ Hence $27(x^2y+y^2z+z^2x+xyz)\leq 27y(x+z)^2=4\cdot27y\cdot\dfrac{x+z}{2}\cdot\dfrac{x+z}{2}\leq 4\left (y+\dfrac{x+z}{2}+\dfrac{x+z}{2}\right )^3=4(x+y+z)^3.$
We are done!

Remark. The above solution was submitted by the AoPS user hoanglong2k $.$

This post has been edited 3 times. Last edited by henderson, Sep 11, 2016, 5:44 PM

Math Path

Problem 29: Useful inequality by Vasile Cirtoaje

by henderson, Jun 29, 2016, 8:43 PM

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