# An Inequality with Constraint III

### Statement

Leo Giugiuc has kindly posted a problem of proving an inequality at the CutTheKnotMath facebook page with a solution by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc and a reference to the *Mathematical Inequalities* facebook group.

Assume $x,y,z$ positive real numbers that satisfy $x^3+y^3+z^3=3.$ Prove that

$\displaystyle \frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\ge 3.$

### Solution

By the Cauchy-Schwarz inequality

$\displaystyle\begin{align} \left(\sum_{cycl}\frac{x^3}{y^2}\right)\left(\sum_{cycl}x^3y^2\right)&\ge\left[\sum_{cycl}\left(\frac{x^{\frac{3}{2}}}{y}\right)\left(x^{\frac{3}{2}}y\right)\right]^2\\ &=\left[\sum_{cycl}x^3\right]^2=9. \end{align}$

Thus the required inequality is equivalent to $\displaystyle\sum_{cycl}\frac{x^3}{y^2}\ge 9\left(\sum_{cycl}x^3y^2\right)^{-1}\;$ which would follow from $\displaystyle\sum_{cycl}x^3y^2\le 3.$

Introduce $a=x^3,\;b=y^3,\;c=z^3.\;$ Then $a+b+c=3\;$ and we need to show that $ab^{\frac{2}{3}}+bc^{\frac{2}{3}}+ca^{\frac{2}{3}}\le 3.\;$ By Jensen's inequality for the concave function $f(t)=t^{\frac{2}{3}},$

$\displaystyle \frac{1}{3}(ab^{\frac{2}{3}}+bc^{\frac{2}{3}}+ca^{\frac{2}{3}})\le\left(\frac{1}{3}(ab+bc+ca)\right)^{\frac{2}{3}}\le 1,$

because $a^2+b^2+c^2\ge ab+bc+ca\;$ and, consequently,

$\displaystyle 9=\left(\sum_{cycl}a\right)^2=\sum_{cycl}a^2+2\sum_{cycl}ab\ge 3\sum_{cycl}ab.$

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