设 $a , b , c>0$，$a+b+c=a b c$．求证：$\displaystyle\sum_{\rm cyc}\dfrac 1{\sqrt{1+a^2}}\leqslant\dfrac 3 2$．
解析 根据题意，有\[\dfrac1{ab}+\dfrac1{bc}+\dfrac1{ca}=1,\]于是\[ \begin{split}\sum_{\rm cyc}\dfrac 1{\sqrt{1+a^2}}&=\sum_{\rm cyc}\dfrac {\dfrac 1a}{\sqrt{\dfrac1{a^2}+\dfrac1{ab}+\dfrac1{bc}+\dfrac1{ca}}}\\ &=\sum_{\rm cyc}\dfrac{\dfrac 1a}{\sqrt{\left(\dfrac 1a+\dfrac 1b\right)\left(\dfrac 1a+\dfrac 1c\right)}}\\ &=\sum_{\rm cyc}\sqrt{\dfrac{\frac 1a}{\frac 1a+\frac 1b}\cdot \dfrac{\frac 1a}{\frac 1a+\frac 1c}}\\ &\leqslant \dfrac 12\sum_{\rm cyc}\left(\dfrac{\frac 1a}{\frac 1a+\frac 1b}+\dfrac{\frac 1a}{\frac 1a+\frac 1c}\right)\\ &=\dfrac 32,\end{split}\]命题得证．