49th Mathematical Olympiad in Ukraine

Denote the length of $AO$ by $x$, $BO$ by $y$ and $CO$ by $z$. From the cosine law we have: (fig.26)

Fig.26

$$AB=\sqrt{x^2+xy+y^2},BC=\sqrt{y^2+yz+z^2},$$ $CA=\sqrt{z^2+zx+x^2}$ and so we can rewrite our inequality: $$\frac{x^2}{\sqrt{y^2+yz+z^2}}+\frac{y^2}{\sqrt{z^2+zx+x^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}\ge \frac{x+y+z}{\sqrt{3}}(1)$$

Without loss of generality we can suppose that $x\ge y\ge z$. Then it is easy to see that $$\frac{1}{\sqrt{y^2+yz+z^2}}\ge \frac{1}{\sqrt{x^2+xz+z^2}}\ge \frac{1}{\sqrt{x^2+xy+y^2}}.\text{ And also it is obvious that }{x}^2\ge y^2\ge z^2.\text{ Thus}$$ we can use an obvious inequality: $$\begin{gathered} \frac{x^2}{\sqrt{y^2+yz+z^2}}+\frac{y^2}{\sqrt{x^2+xz+z^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}\ge \frac{y^2}{\sqrt{y^2+yz+z^2}}+\frac{z^2}{\sqrt{x^2+xz+z^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}, \\ \frac{x^2}{\sqrt{y^2+yz+z^2}}+\frac{y^2}{\sqrt{x^2+xz+z^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}\ge \frac{z^2}{\sqrt{y^2+yz+z^2}}+\frac{x^2}{\sqrt{x^2+xz+z^2}}+\frac{y^2}{\sqrt{x^2+xy+y^2}} \end{gathered}$$ Therefore the left hand side of the inequality (1) is no less than $$\frac{1}{2}\left(\frac{y^2+z^2}{\sqrt{y^2+yz+z^2}}+\frac{x^2+z^2}{\sqrt{x^2+xz+z^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}\right).$$ We will now show that $\frac{y^2+z^2}{\sqrt{y^2+yz+z^2}}\ge \frac{y+z}{\sqrt{3}}$. To prove this let us use the power mean inequality: $$\frac{y^2+z^2}{\sqrt{y^2+yz+z^2}}\ge \sqrt{\frac{2}{3}(y^2+z^2)}=\frac{2}{\sqrt{3}}\sqrt{\frac{y^2+z^2}{2}}\ge \frac{2}{\sqrt{3}}\cdot \frac{y+z}{2}=\frac{y+z}{\sqrt{3}}$$ In the same way we can get analogous inequalities for pairs $(x,y)$ and $(x,z)$. And so we finally get: $$\frac{y^2+z^2}{\sqrt{y^2+z^2}}+\frac{x^2+z^2}{\sqrt{x^2+xz+z^2}}+\frac{x^2+y^2}{\sqrt{x^2+xy+y^2}}\ge \frac{2(x+y+z)}{\sqrt{3}},$$ which was to be proved.

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Alternative solution.

Here we will give another proof of the inequality (1). Using obvious inequalities $yz\le \frac{y^2+z^2}{2}$, $xz\le \frac{x^2+z^2}{2}$ and $xy\le \frac{x^2+y^2}{2}$ we can obtain: $$\frac{x^2}{\sqrt{y^2+yz+z^2}}+\frac{y^2}{\sqrt{z^2+zx+x^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}\ge \sqrt{\frac{2}{3}}\left(\frac{x^2}{\sqrt{y^2+z^2}}+\frac{y^2}{\sqrt{z^2+y^2}}+\frac{z^2}{\sqrt{x^2+y^2}}\right).$$ Now denote $S=x^2+y^2+z^2$ and consider the function $f(t)=\frac{1}{\sqrt{S-t}}$. Since $$f^{\prime \prime }(t)=\frac{1}{\sqrt{(S-t{)}^3}}+\frac{3}{4}\frac{1}{\sqrt{(S-t{)}^5}}=\frac{4(S-t)+3t}{4\sqrt{(S-t{)}^5}}=\frac{4S-t}{4\sqrt{(S-t{)}^5}}\ge 0$$ for $0\le t<S$, function $f$ is convex on $[0,S)$. And so applying the Jensen's inequality for $(0\le x^2,y^2,z^2<S$ we can get: $$\begin{gathered} \frac{1}{3}\left(\frac{x^2}{\sqrt{y^2+z^2}}+\frac{y^2}{\sqrt{z^2+x^2}}+\frac{z^2}{\sqrt{x^2+y^2}}\right)=\frac{1}{3}\left(\frac{x^2}{\sqrt{S-x^2}}+\frac{y^2}{\sqrt{S-y^2}}+\frac{z^2}{\sqrt{S-z^2}}\right)= \\ =\frac{1}{3}\left(f(x^2)+f(y^2)+f(z^2)\right)\ge f\left(\frac{x^2+y^2+z^2}{3}\right)=\frac{8}{3\sqrt{S-\frac{8}{3}}}=\sqrt{\frac{(x^2+y^2+z^2{)}^2}{6}}\ge \frac{x+y+z}{3\sqrt{2}} \end{gathered}$$ And finally $$\frac{x^2}{\sqrt{y^2+yz+z^2}}+\frac{y^2}{\sqrt{z^2+zx+x^2}}+\frac{z^2}{\sqrt{x^2+xy+y^2}}\ge \sqrt{\frac{2}{3}}\cdot \frac{1}{\sqrt{2}}(x+y+z)=\frac{x+y+z}{\sqrt{3}},$$ And we're done.