Does anybody have solutions?

This problem was Turkmenistan National Math Olympaid 2013.

vi1lat wrote: If a,b,c positive numbers and such that $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}$. Prove that if $a\neq c$ then $40ac<1$. $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}\Leftrightarrow a-c=\frac{\sqrt a-\sqrt c}{\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}}\Leftrightarrow$ $\Leftrightarrow\left(%Error. "sqty" is a bad command. a+\sqrt c\right)\left(\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}\right)=1$. Hence, $1=\left(\sqrt a+\sqrt c\right)\left(\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}\right)\geq\left(\sqrt a+\sqrt c\right)\left(\sqrt{\sqrt{a}}+\sqrt{\sqrt{c}}\right)\geq$ $\geq2\sqrt[4]{ac}\cdot2\sqrt[8]{ac}=\sqrt[8]{2^{16}a^3c^3}=\left(2^{\frac{16}{3}}ac\right)^{\frac{3}{8}}>(40ac)^{\frac{3}{8}}$. Id est, $40ac<1$.

arqady wrote: vi1lat wrote: If a,b,c positive numbers and such that $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}$. Prove that if $a\neq c$ then $40ac<1$. $a+\sqrt{b+\sqrt{c}}=c+\sqrt{b+\sqrt{a}}\Leftrightarrow a-c=\frac{\sqrt a-\sqrt c}{\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}}\Leftrightarrow$ $\Leftrightarrow\left(%Error. "sqty" is a bad command. a+\sqrt c\right)\left(\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}\right)=1$. Hence, $1=\left(\sqrt a+\sqrt c\right)\left(\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}\right)\geq\left(\sqrt a+\sqrt c\right)\left(\sqrt{\sqrt{a}}+\sqrt{\sqrt{c}}\right)\geq$ $\geq2\sqrt[4]{ac}\cdot2\sqrt[8]{ac}=\sqrt[8]{2^{16}a^3c^3}=\left(2^{\frac{16}{3}}ac\right)^{\frac{3}{8}}>(40ac)^{\frac{3}{8}}$. Id est, $40ac<1$. Nice. $1=\left(\sqrt a+\sqrt c\right)\left(\sqrt{b+\sqrt{a}}+\sqrt{b+\sqrt{c}}\right)>\left(\sqrt a+\sqrt c\right)\left(\sqrt{\sqrt{a}}+\sqrt{\sqrt{c}}\right)$... It is Turkmenistan National Math Olympiad 2013 Problem 3