Let $AO = a, BO=b, CO=c$ By the cosine law we have that: ${AB = \sqrt{a^2+b^2+ab}, BC = \sqrt{b^2+c^2+bc}, CA=\sqrt c^2+a^2+ca}$ Then we need to prove that: $\frac{a^2}{\sqrt{b^2+c^2+bc}} + \frac{b^2}{\sqrt{c^2+a^2+ca}} + \frac{c^2}{\sqrt{a^2+b^2+ab}} \ge \frac{a+b+c}{\sqrt 3}$ This inequality is homogeneous, so we can assume $a^2+b^2+c^2=1$ We know that the function $f(x) = \frac 1\sqrt{x}$ is convex for $x\in (0,\infty)$. $\frac{a^2}{\sqrt{b^2+c^2+bc}} + \frac{b^2}{\sqrt{c^2+a^2+ca}} + \frac{c^2}{\sqrt{a^2+b^2+ab}}$ $\ge \frac 1{\sqrt{a^2(b^2+c^2+bc)+b^2(c^2+a^2+ca)+c^2(a^2+b^2+ab)}}$ by Jensen $= \frac 1{\sqrt{a^2bc+b^2ca+c^2ab+2a^2b^2+2b^2c^2+2c^2a^2}}$ $\ge \frac 1{\sqrt{a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2}}$ since $a^4+b^4+c^4 \ge a^2bc+b^2ca+c^2ab$ by Muirhead $= \frac 1{a^2+b^2+c^2} = 1 = \sqrt{a^2+b^2+c^2} \ge \frac{a+b+c}{\sqrt{3}}$ by Cauchy Schwarz

m.candales wrote: . $\frac{a^2}{\sqrt{b^2+c^2+bc}} + \frac{b^2}{\sqrt{c^2+a^2+ca}} + \frac{c^2}{\sqrt{a^2+b^2+ab}}$ $\ge \frac 1{\sqrt{a^2(b^2+c^2+bc)+b^2(c^2+a^2+ca)+c^2(a^2+b^2+ab)}}$ by Jensen Don't see how? thanks in advance

For $f$ a convex function, $\alpha f(x) + \beta f(y) + \gamma f(z) \geq f(\alpha x + \beta y + \gamma z)$, for all $x,y,z$ in the (convex) domain of $f$, and all non-negative real numbers $\alpha, \beta, \gamma$ with $\alpha + \beta + \gamma = 1$. Take $f(x) = \dfrac {1} {\sqrt{x}}$, $\alpha = a^2$, $\beta = b^2$, $\gamma = c^2$, $x = b^2 + c^2 + bc$, $y = c^2 + a^2 + ca$, $z = a^2 + b^2 + ab$.

who can solve it by cauchy inequality.

chinacai wrote: who can solve it by cauchy inequality. I think I can. We rewrite the problem into, letting $a=AO, b=BO, c=CO;$ \[\frac{a^2}{\sqrt{b^2+bc+c^2}}+\frac{b^2}{\sqrt{c^2+ca+a^2}}+\frac{c^2}{\sqrt{a^2+ab+b^2}}\geq \frac{a+b+c}{\sqrt 3}.\] Note that using the Holder inequality we have \[\sum_{cyc}\frac{a^3}{a\sqrt{b^2+bc+c^2}}\geq \frac{(a+b+c)^3}{3\sum a\sqrt{b^2+bc+c^2}}.\] Now we have, using the Cauchy-Schwarz inequality that \[\begin{aligned}\sum_{cyc}a\sqrt{b^2+bc+c^2}&\leq \sqrt{(a+b+c)\sum a(b^2+bc+c^2)}\\&=(a+b+c)\sqrt{ab+bc+ca};\end{aligned}\] And therefore it is sufficient to show that $(a+b+c)\geq \sqrt{3(ab+bc+ca)};$ which is an obvious consequence of the AM-GM inequality. Equality holds iff $O$ is the circumcentre of $\triangle ABC\iff \angle A=\angle B=\angle C=\dfrac{\pi}{3}.\Box$

holder inequality

Attachments:
UKMO2009.pdf (60kb)

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=267383&hilit=Ukraine+inequality