# Inequality 47(Christos Patilas)

Problem:

If $\displaystyle x_i$ for $\displaystyle i=1,2,...,n$ are positive real numbers then prove that

$\displaystyle \sum^{n}_{i=1}\left(5\sqrt[5]{x^{3}_{i}}-3\sqrt[3]{\left(\frac{3x_i+2}{5}\right)^5}\right)\leq 2n$

Solution(An Idea by Vo Quoc Ba Can):

We only need to prove that

$\displaystyle 5\sqrt[5]{a^3}-3\sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\leq 2$ for all $\displaystyle a>0$.

So, using the AM-GM Inequality we have that

$\displaystyle a+a+a+1+1\geq 5\sqrt[5]{a^3}$.

It follows that

$\displaystyle \sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\geq \sqrt[3]{\left(\sqrt[5]{a^3}\right)^5}=a$.

Therefore it suffices to prove that

$\displaystyle 5\sqrt[5]{a^3}-2\leq 3a$,

which is obviously true from the AM-GM Inequality, Q.E.D.