Inequality 3

by MM » Wed Sep 10, 2008 12:51 pm

If $a,b,c\ge 0$ and $a+b+c=\frac{3}{\sqrt{2}}$ , prove that:
$\frac{a^{3}+b^{3}+c^{3}}{3}+\frac{\sqrt{5\left(a^{2}+b^{2}+c^{2}\right)}}{\sqrt{6}}+abc+\frac{2\sqrt{2}-\sqrt{5}}{2}\ge a\sqrt[3]{ab+ac}+b\sqrt[3]{ba+bc}+c\sqrt[3]{ca+cb}$ .

Proposed by me

by **martosss** » Wed Sep 17, 2008 10:47 am

MM wrote:If $a,b,c\ge 0$ and $a+b+c=\frac{3}{\sqrt{2}}$ , prove that:
$\frac{a^{3}+b^{3}+c^{3}}{3}+\frac{\sqrt{5\left(a^{2}+b^{2}+c^{2}\right)}}{\sqrt{6}}+abc+\frac{2\sqrt{2}-\sqrt{5}}{2}\ge a\sqrt[3]{ab+ac}+b\sqrt[3]{ba+bc}+c\sqrt[3]{ca+cb}$ .

Proposed by me

First we will prove, that $\frac{\sqrt{5(a^2+b^2+c^2}}{\sqrt{6}}\ge \frac{\sqrt{5}}{2}$ . That is equivalent to $a^2+b^2+c^2\ge \frac{3}{2}$ , which is RMS≥AM

Furthermore from AM≥GM we have

$\frac{(ab+ac)+1+1}{3}\ge \sqrt[3]{ab+bc}$

$\frac{(ba+bc)+1+1}{3}\ge \sqrt[3]{ba+bc}$

$\frac{(ac+bc)+1+1}{3}\ge \sqrt[3]{ac+bc}$

Now let us multiply (1), (2), and (3) respectively by a, b and c, after that let us add the 3 equations - we get:

$\frac{a(ab+ac)+2a}{3}+\frac{b(ab+bc)+2b}{3}+\frac{c(ac+bc)+2c}{3}\ge a\sqrt[3]{ab+ac}+ b\sqrt[3]{ab+bc} +c\sqrt[3]{ac+bc}$

Now we must prove, that

$\frac{a^{3}+b^{3}+c^{3}+3abc+\overbrace{3\sqrt{2}}^{=\sqrt{2}(a+b+c)\sqrt{2}=2(a+b+c)}}{3}\ge \frac{a(ab+ac)+2a}{3}+\frac{b(ab+bc)+2b}{3}+\frac{c(ac+bc)+2c}{3}\Leftright$

$\Leftright a^3+b^3+c^3+3abc\ge a(ab+ac)+b(ab+bc)+c(ac+bc)$

which is equivalent to Schur's inequality.

by MM » Wed Sep 17, 2008 1:13 pm

Good work martosss!

Inequality 3

Inequality 3

Re: Inequality 3

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