求证：对任意正实数 $a、b、c$，都有 $1<\dfrac{a}{\sqrt{a^{2}+b^{2}}}+\dfrac{b}{\sqrt{b^{2}+c^{2}}}+\dfrac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \dfrac{3 \sqrt{2}}{2}$．
【难度】
【出处】
2004年中国西部数学奥林匹克试题
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【答案】
略
【解析】
令 $x=\dfrac{b^{2}}{a^{2}}, y=\dfrac{c^{2}}{b^{2}}, z=\dfrac{a^{2}}{c^{2}}$，则 $x, y, z \in \mathbf{R}^{+}, x y z=1$．于是只需证明 $1<\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \leqslant \dfrac{3 \sqrt{2}}{2}$．
不妨设 $0<x \leqslant y \leqslant z$，令 $A=x y$，则 $z=\dfrac{1}{A}, A \leqslant 1$．于是
$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}}>\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+\frac{1}{x}}}=\dfrac{1+\sqrt{x}}{\sqrt{1+x}}>1$．
设 $u=\dfrac{1}{\sqrt{1+A+x+\frac{A}{x}}}$，则 $u \in\left(0, \dfrac{1}{1+\sqrt{A}}\right]$，当且仅当 $x=\sqrt{A}$ 时，$u=\dfrac{1}{1+\sqrt{A}}$．于是
$\begin{aligned}
&\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\right)^{2}\\
=&\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+\frac{A}{x}}}\right)^{2} \\
=&\frac{1}{1+x}+\frac{1}{1+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\
=&\frac{2+x+\frac{A}{x}}{1+A+x+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\
=&1+(1-A) u^{2}+2 u \end{aligned}$
令 $f(u)=(1-A) u^{2}+2 u+1$，则 $f(u)$ 在 $u \in\left(0, \frac{1}{1+\sqrt{A}}\right]$ 上是增函数，所以 $\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}} \leqslant \sqrt{f\left(\dfrac{1}{1+\sqrt{A}}\right)}=\dfrac{2}{\sqrt{1+\sqrt{A}}}$．
令 $\sqrt{A}=v$，则
$\begin{aligned} & \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \leqslant \frac{2}{\sqrt{1+\sqrt{A}}}+\frac{1}{\sqrt{1+\frac{1}{A}}} \\=& \frac{2}{\sqrt{1+v}}+\frac{\sqrt{2} v}{\sqrt{2\left(1+v^{2}\right)}} \leqslant \frac{2}{\sqrt{1+v}}+\frac{\sqrt{2} v}{1+v} \\=& \frac{2}{\sqrt{1+v}}+\sqrt{2}-\frac{\sqrt{2}}{1+v} \\=&-\sqrt{2}\left(\frac{1}{\sqrt{1+v}}-\frac{\sqrt{2}}{2}\right)^{2}+\frac{3 \sqrt{2}}{2} \\ \leqslant & \frac{3 \sqrt{2}}{2} \end{aligned}$
不妨设 $0<x \leqslant y \leqslant z$，令 $A=x y$，则 $z=\dfrac{1}{A}, A \leqslant 1$．于是
$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}}>\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+\frac{1}{x}}}=\dfrac{1+\sqrt{x}}{\sqrt{1+x}}>1$．
设 $u=\dfrac{1}{\sqrt{1+A+x+\frac{A}{x}}}$，则 $u \in\left(0, \dfrac{1}{1+\sqrt{A}}\right]$，当且仅当 $x=\sqrt{A}$ 时，$u=\dfrac{1}{1+\sqrt{A}}$．于是
$\begin{aligned}
&\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\right)^{2}\\
=&\left(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+\frac{A}{x}}}\right)^{2} \\
=&\frac{1}{1+x}+\frac{1}{1+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\
=&\frac{2+x+\frac{A}{x}}{1+A+x+\frac{A}{x}}+\frac{2}{\sqrt{1+A+x+\frac{A}{x}}} \\
=&1+(1-A) u^{2}+2 u \end{aligned}$
令 $f(u)=(1-A) u^{2}+2 u+1$，则 $f(u)$ 在 $u \in\left(0, \frac{1}{1+\sqrt{A}}\right]$ 上是增函数，所以 $\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}} \leqslant \sqrt{f\left(\dfrac{1}{1+\sqrt{A}}\right)}=\dfrac{2}{\sqrt{1+\sqrt{A}}}$．
令 $\sqrt{A}=v$，则
$\begin{aligned} & \frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}} \leqslant \frac{2}{\sqrt{1+\sqrt{A}}}+\frac{1}{\sqrt{1+\frac{1}{A}}} \\=& \frac{2}{\sqrt{1+v}}+\frac{\sqrt{2} v}{\sqrt{2\left(1+v^{2}\right)}} \leqslant \frac{2}{\sqrt{1+v}}+\frac{\sqrt{2} v}{1+v} \\=& \frac{2}{\sqrt{1+v}}+\sqrt{2}-\frac{\sqrt{2}}{1+v} \\=&-\sqrt{2}\left(\frac{1}{\sqrt{1+v}}-\frac{\sqrt{2}}{2}\right)^{2}+\frac{3 \sqrt{2}}{2} \\ \leqslant & \frac{3 \sqrt{2}}{2} \end{aligned}$
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