Home » Uncategorized » Inequality 12(Vasile Cirtoaje)

# Inequality 12(Vasile Cirtoaje)

Problem:

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

$\displaystyle (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)\geq (ab+bc+ca)^3$.

Solution:

Lemma: $\displaystyle 4(a^2+ab+b^2)\geq 3(a+b)^2$.

Back to the inequality now, multiply both sides by $\displaystyle 64$. Then we have that $\displaystyle 4^3\prod_{cyc}(a^2+ab+b^2)\geq 4^3(ab+bc+ca)^3$.

But from the lemma we reduce the current inequality to

$\displaystyle 27\prod_{cyc}(a+b)^2\geq 64(ab+bc+ca)^3$.

It also holds $\displaystyle (a+b+c)^2\geq 3(ab+bc+ca)$. Multiplying the last inequality with $\displaystyle \frac{64}{3}(ab+bc+ca)^2$ we get that $\displaystyle \frac{64}{3}(ab+bc+ca)^2(a+b+c)^2\geq 64(ab+bc+ca)^3$.

So, it suffices to prove that

$\displaystyle 27\prod_{cyc}(a+b)^2\geq \frac{64}{3}(a+b+c)^2(ab+bc+ca)^2$

or

$\displaystyle 9(a+b)(b+c)(c+a)\geq 8(a+b+c)(ab+bc+ca)$,

which reduces to the obvious inequality

$\displaystyle a(b-c)^2+b(c-a)^2+c(a-b)^2\geq 0$.

Equality occurs for $\displaystyle (a,b,c)=(1,1,1)$ and also for $\displaystyle (a,b,c)=(1,0,0)$ or any cyclic permutation, Q.E.D.