45th Mongolian Mathematical Olympiad

First, we will prove the following inequality. If $a,b,c$ are positive real numbers then $$\frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\ge 1.(1)$$ Indeed, we need to find $\lambda \in \mathbb{R}$ such that $$\frac{a}{\sqrt{a^2+bc}}\ge \frac{a^{\lambda }}{b^{\lambda }+c^{\lambda }+a^{\lambda }}.(2)$$ From (2), we get $$a^2(a^{\lambda }+b^{\lambda }+c^{\lambda }{)}^2\ge a^{2\lambda }(a^2+bc).(3)$$ $$\begin{array}{rl}{a}^2(a^{\lambda }+b^{\lambda }+c^{\lambda }{)}^2& =a^{2\lambda +2}+a^2(b^{2\lambda }+2a^2{b}^{\lambda }{c}^{\lambda }+2a^{\lambda +2}{b}^{\lambda }+2a^{\lambda +2}{c}^{\lambda })\\ & \ge a^{2\lambda +2}+2a^2{b}^{\lambda }{c}^{\lambda }+2a^2{b}^{\lambda }{c}^{\lambda }+2a^{\lambda +2}{b}^{\lambda }+2a^{\lambda +2}{c}^{\lambda }\\ & \ge a^{2\lambda +2}+4\sqrt[4]{2^4{a}^{2\lambda +8}{b}^{3\lambda }{c}^{3\lambda }}=a^{2\lambda +2}+8a^{\frac{\lambda +4}{2}}{b}^{\frac{3\lambda }{4}}{c}^{\frac{3\lambda }{4}}.\end{array}(4)$$ By (3) and (4), we get $$a^{2\lambda +2}+8a^{\frac{\lambda +4}{2}}{b}^{\frac{3\lambda }{4}}{c}^{\frac{3\lambda }{4}}=a^{2\lambda +2}+8a^{2\lambda }bc.(5)$$ From (5), by easy calculation we find that $\lambda =\frac{4}{3}$. Now from (2) by easy way get (1). In the (1) inequality substituting $\frac{a^2}{bc}=x$, $\frac{b^2}{ac}=y$, $\frac{c^2}{ab}=z$ then $xyz=1$, $x,y,z>0$; we get $$\sqrt{\frac{x}{x+8}}+\sqrt{\frac{y}{y+8}}+\sqrt{\frac{z}{z+8}}\ge 1.(6)$$ Let $(x^{\prime },y^{\prime },z^{\prime })$ be an arbitrary permutation of $(x,y,z)$. The following triples are inversely monotonic $(\sqrt{x};\sqrt{y};\sqrt{z})$ and $(\frac{1}{\sqrt{x+8}},\frac{1}{\sqrt{y+8}},\frac{1}{\sqrt{z+8}})$ hence by rearrangement method, one gets $$\sqrt{\frac{x}{x+8}}+\sqrt{\frac{y}{y+8}}+\sqrt{\frac{z}{z+8}}\le \sqrt{\frac{x^{\prime }}{y^{\prime }+8}}+\sqrt{\frac{y^{\prime }}{z^{\prime }+8}}+\sqrt{\frac{z^{\prime }}{x^{\prime }+8}}$$ Now taking this $(x,y,z)=(a,b,c)$, $x^{\prime },y^{\prime },z^{\prime }=(a,b,c)$ then our inequality is proved. Equality holds for $a=b=c=1$.