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Inequality 12(Vasile Cirtoaje)


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.


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


\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.


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