Russian inequality

Cekaa · New Member Offline Joined: 13 Apr 2008 Posts: 19

Prove that for all positive reals $a,b,c$ :
$\frac{a^2b+b^2c+c^2a}{a^3+b^3+c^3}+\frac{3(a^2+b^2+c^2)}{ab+bc+ca} \geq 4$

**Posted:** Fri Nov 06, 2009 10:45 am

Abdek

We have the flowing identities:

$\left(a^2+b^2+c^2-ab-bc-ca\right)=\frac{1}{2}\sum_{cyc}(a-b)^2$
$\left(a^3+b^3+c^3-a^2b-b^2c-c^2a\right)=\frac{1}{3}\sum_{cyc}(2a+b)(a-b)^2$

Hence:

$\text{LHS} - 4 = \frac {3(a^2 + b^2 + c^2 - ab - bc - ca)}{ab + bc + ca} + \frac {a^2b + b^2c + c^2a - a^3 - b^3 - c^3}{a^3 +...$
$= \sum_{cyc}\frac {3(a - b)^2}{2(ab + bc + ca)} - \sum_{cyc}\frac {(a - b)^2(2a + b)}{3(a^3 + b^3 + c^3)}$
$= \sum_{cyc}\frac {(a - b)^2(9a^3 + 9b^3 + 9c^3 - 4a^2c - 4a^2b - 4abc - a^2c - a^2b - abc)}{6(ab + bc + ca)(a^3 + b^3 + c^3)...$

which is true ! Embarassed

**Posted:** Fri Nov 06, 2009 11:18 am

_________________
Mharchi Abdelmalek

arqady

**Posted:** Fri Nov 06, 2009 1:16 pm

_________________
Michael Rozenberg

Dumel

in my solution i used $(a^2b+b^2c+c^2a)(b+c+a) \ge (ab+bc+ca)^2$ and then p,q,r works

**Posted:** Sat Nov 07, 2009 9:26 am