Home » Uncategorized » Inequality 4(Vasile Cirtoaje)

# Inequality 4(Vasile Cirtoaje)

Problem:

If $\displaystyle a,b,c$ are non-negative numbers prove that

$\displaystyle 2(a^2+1)(b^2+1)(c^2+1)\geq (a+1)(b+1)(c+1)(abc+1)$.

Solution:

For $\displaystyle a=b=c$ the inequality reduces to

$\displaystyle 2(a^2+1)^{3}\geq (a+1)^{3}(a^3+1)$.

This inequality is true since

$\displaystyle 2(a^2+1)^{3}-(a+1)^{3}(a^3+1)=(a-1)^{4}(a^2+a+1)\geq 0$.

Multiplying now the inequalities:

$\displaystyle 2(a^2+1)^{3}\geq (a+1)^{3}(a^3+1)$,

$\displaystyle 2(b^2+1)^{3}\geq (b+1)^{3}(b^3+1)$,

$\displaystyle 2(c^2+1)^{3}\geq (c+1)^{3}(c^3+1)$,

we get $\displaystyle 8\prod_{cyc}(a^2+1)^{3}\geq \prod_{cyc}(a+1)^{3}(a^3+1)$.

So, we still have to show that

$\displaystyle \prod_{cyc}(a^3+1)\geq (abc+1)^{3}$.

But the last inequality is true since from Holder’s inequality we have that

$\displaystyle \prod_{cyc}(a^3+1)\geq \left(\sqrt[3]{a^3b^3c^3}+\sqrt[3]{1\cdot 1\cdot 1}\right)^{3}=(abc+1)^{3}$, Q.E.D.

Advertisements