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# Inequality 21(Vasile Cirtoaje)

Problem:

Let $\displaystyle x,y,z$ be non-negative numbers. If $\displaystyle 0\leq r\leq \sqrt{2}$, prove that

$\displaystyle \sqrt{x^4+y^4+z^4}+r\sqrt{x^2y^2+y^2z^2+z^2x^2}\geq (1+r)\sqrt{x^3y+y^3z+z^3x}$.

Solution:

If we square both sides we get

$\displaystyle \sum_{cyc}x^4+r^2\sum_{cyc}x^2y^2+2r\sqrt{\sum_{cyc}x^4\sum_{cyc}x^2y^2}\geq r^2\sum_{cyc}x^3y+\sum_{cyc}x^3y+2r\sum_{cyc}x^3y$.

Now from Cauchy-Schwartz inequality we know that $\displaystyle 2r\sqrt{\sum_{cyc}x^4\sum_{cyc}x^2y^2}\geq 2r\sum_{cyc}x^3y$.

So, it suffices to prove that

$\displaystyle \sum_{cyc}x^4+r^2\sum_{cyc}x^2y^2\geq (1+r^2)\sum_{cyc}x^3y$.

For $\displaystyle r=0$ the inequality is true. So, we only need to prove it for $\displaystyle 0.

Rewrite the inequality in the form

$\displaystyle \sum_{cyc}x^4-\sum_{cyc}x^3y\geq r^2\left(\sum_{cyc}x^3y-\sum_{cyc}x^2y^2\right)$.

We know that $\displaystyle \sum_{cyc}x^4-\sum_{cyc}x^3y\geq 0$ so, it is enough to prove it for $\displaystyle r=\sqrt{2}$.

For $\displaystyle r=\sqrt{2}$ we have that

$\displaystyle \sum_{cyc}x^4-\sum_{cyc}x^3y\geq 2\left(\sum_{cyc}x^3y-\sum_{cyc}x^2y^2\right)$,

which reduces to

$\displaystyle \left(\sum_{cyc}x^2\right)^2\geq 3\sum_{cyc}x^3y$, which is a well known inequality and we have proved it here: Well-known inequality, Q.E.D.