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Inequality 4(Vasile Cirtoaje)


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).


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.


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