Inequality 2

by MM » Wed Sep 10, 2008 9:29 am

If $a,b,c\ge 0$ and $a+b+c=\frac{3}{2}$ prove that:
$4\left(a^{3}+b^{3}+c^{3}\right)+12abc+9\ge 8\left(a^{2}\sqrt{b+c}+b^{2}\sqrt{c+a}+c^{2}\sqrt{a+b}+ab+bc+ca\right)$

I am the author.

by **martosss** » Mon Sep 22, 2008 8:21 am

Let us add $4(a^2+b^2+c^2)$ to both sides of the inequality:

$4(a^3+b^3+c^3)+12abc+4(a^2+b^2+c^2)+\N 9\ge 8(a^2\sqrt{b+c}+b^2\sqrt{a+c}+c^2\sqrt{a+b})+\N {\underbrace{4(a+b+c)^2}_{=9}}$

Now from AM≥GM we have

$\frac{(b+c)+1}{2}\ge\sqrt{b+c}$

$\frac{\(a+c)+1}{2}\ge\sqrt{a+c}$

$\frac{(a+b)+1}{2}\ge\sqrt{a+b}$

Now let us multiply (1), (2) and (3) respectively with a², b² and c², then add the 3 inequalities and multiply that by 8 - we get

$4\left[a^2(b+c)+b^2(a+c)+c^2(a+b)+a^2+b^2+c^2\right]\ge 8(a^2\sqrt{b+c}+b^2\sqrt{a+c}+c^2\sqrt{a+b})$

So we must prove, that

$\N 4(a^3+b^3+c^3)+\overbrace{\N {12}}^{3}abc+\N{4(a^2+b^2+c^2)}\ge \N 4\left[a^2(b+c)+b^2(a+c)+c^2(a+b)\right]+\N {4\left[a^2+b^2+c^2\right]}$

That leaves the Schur's Inequality and we're done :wink:

Inequality 2

Inequality 2

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