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Inequality 44(Unknown Author)

Problem:

Let \displaystyle a,b,c be positive real numbers. Prove that

\displaystyle 1+\frac{8abc}{(a+b)(b+c)(c+a)}\geq \frac{2(ab+bc+ca)}{a^2+b^2+c^2}.

Solution (An idea by Silouanos Brazitikos):

From the above inequality is it enough to show that

\displaystyle \frac{8abc}{(a+b)(b+c)(c+a)}\geq \frac{2(ab+bc+ca)-a^2+b^2+c^2}{a^2+b^2+c^2}.

But from Schur’s Inequality we know that

\displaystyle 2(ab+bc+ca)-a^2-b^2-c^2\leq \frac{9abc}{a+b+c}.

So it is enough to check that

\displaystyle 8(a^2+b^2+c^2)(a+b+c)\geq 9(a+b)(b+c)(c+a).

From Cauchy-Schwarz Inequality we know that

\displaystyle 3(a^2+b^2+c^2)\geq (a+b+c)^2.

Therefore we only need to prove that

\displaystyle \left(\frac{a+b+b+c+c+a}{3}\right)^{3}\geq (a+b)(b+c)(c+a),

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

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