每日一题[1420]追求轮换

已知 $a,b,c>0$，且 $abc=1$，求证：$\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\leqslant \dfrac 12$．

解析    令 $(a,b,c)=\left(\dfrac yz,\dfrac zx,\dfrac xy\right)$，$x,y,z>0$，则

$$\begin{split} LHS&=\sum_{\rm cyc}\dfrac{1}{\dfrac{y^2}{z^2}+\dfrac{2z^2}{x^2}+3}\\ &=\sum_{\rm cyc}\dfrac{1}{\left(\dfrac {y^2}{z^2}+\dfrac{z^2}{x^2}\right)+\left(\dfrac{z^2}{x^2}+1\right)+2}\\ &\leqslant \sum_{\rm cyc}\dfrac{1}{\dfrac{2y}{x}+\dfrac{2z}{x}+2}\\ &=\sum_{\rm cyc}\dfrac{x}{2(x+y+z)}\\ &=\dfrac 12,\end{split}$$ 因此命题成立．