问题:
若$x^3+y^3+z^3=3$,且$x,y,z>0$，证明
\[ \frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\leq \frac{1}{2} \]
证明：（西西）
由AM-GM
\[ \frac{x}{y^2+5}=\frac{1}{6}\cdot x\cdot \frac{6}{y^2+5}\leq \frac{1}{6}\left(\frac{x^2+\frac{6}{(y^2+5)^2}}{2}\right)=\frac{x^2}{12}+\frac{3}{(y^2+5)^2} \]
这样只要证明
\[ \sum_{cyc}{\frac{x^2}{12}}+\sum_{cyc}{\frac{3}{(y^2+5)^2}}\leq \frac{1}{2} \]
考虑
\[ f(x)=\frac{x^2}{12}+\frac{3}{(x^2+5)^2} \]
对$x^3$的切线，得到
\[ \frac{x^2}{12}+\frac{3}{(x^2+5)^2}\leq \frac{2x^3+7}{54}\]
\[ \Leftrightarrow 6(x-1)^2(4x^5-x^4+34x^3-7x^2+52x+26)\geq 0 \qquad x\in [0,\sqrt[3]{3}]\]
故有
\[ \frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\leq \frac{2(x^3+y^3+z^3)+21}{54}=\frac{1}{2}\]