China Western Mathematical Olympiad

Set $x=\frac{b^2}{a^2}$, $y=\frac{c^2}{b^2}$, $z=\frac{a^2}{c^2}$, then $x,y,z\in {\mathbb{R}}^{+}$ and $xyz=1$. It suffices to prove that $$1<\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}\le \frac{3\sqrt{2}}{2}.$$

Without loss of generality, we assume that $x\le y\le z$. Set $A=xy$, we have $z=\frac{1}{A}$, $A\le 1$. Thus $$\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}>\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+\frac{1}{x}}}=\frac{1+\sqrt{x}}{\sqrt{1+x}}>1.$$ Let $u=\frac{1}{\sqrt{1+A+x+\frac{A}{x}}}$, then $u\in (0,\frac{1}{1+\sqrt{A}}]$, and $x=\sqrt{A}$ if and only if $u=\frac{1}{1+\sqrt{A}}$. Hence $${\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\right)}^2={\left[\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+\frac{A}{x}}}\right]}^2$$ $$\begin{gathered} =\frac{1}{1+x}+\frac{1}{1+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\ =\frac{2+x+\frac{A}{x}}{1+A+x+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\ =1+(1-A)u^2+2u. \end{gathered}$$ Set $f(u)=(1-A)u^2+2u+1$, we see that $f(u)$ is an increasing function on $u\in (0,\frac{1}{1+\sqrt{A}}]$, which implies that $$\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\le \sqrt{f\left(\frac{1}{1+\sqrt{A}}\right)}=\frac{2}{\sqrt{1+\sqrt{A}}}$$ Now set $\sqrt{A}=v$, we get $$\begin{gathered} \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}\le \frac{2}{\sqrt{1+\sqrt{A}}}+\frac{1}{\sqrt{1+\frac{1}{A}}} \\ =\frac{2}{\sqrt{1+v}}+\frac{\sqrt{2}v}{\sqrt{2}(1+v^2)}\le \frac{2}{\sqrt{1+v}}+\frac{\sqrt{2}v}{1+v} \\ =\frac{2}{\sqrt{1+v}}+\sqrt{2}-\frac{\sqrt{2}}{1+v}=-\sqrt{2}{\left[\frac{1}{\sqrt{1+v}}-\frac{\sqrt{2}}{2}\right]}^2+\frac{3\sqrt{2}}{2} \\ \le \frac{3\sqrt{2}}{2}. \end{gathered}$$