设$a,b,c>0$且$a+b+c=3$,$k>0$,证明
\[ (b+c)\sqrt[k]{\frac{bc+1}{a^2+1}}+(c+a)\sqrt[k]{\frac{ca+1}{b^2+1}}+(a+b)\sqrt[k]{\frac{ab+1}{c^2+1}}\geq 6 \]
证明
不等式等价于
\[ \sum{a\left(\sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\right)}\geq 6 \]
由AM-GM,
\[ k\cdot\sqrt[k]{\frac{ca+1}{b^2+1}}+k\cdot\sqrt[k]{\frac{ab+1}{c^2+1}}+\frac{b^2+1}{ab+1}+\frac{c^2+1}{ca+1}\geq 2(k+1) \]
得到
\[ \sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\geq \frac{2(k+1)}{k}-\frac{1}{k}\left(\frac{b^2+1}{ab+1}+\frac{c^2+1}{ac+1}\right)\]
\[a\left(\sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\right)\geq \frac{2(k+1)a}{k}-\frac{1}{k}\left(\frac{a(b^2+1)}{ab+1}+\frac{a(c^2+1)}{ac+1}\right)\]
\begin{align*}
 \sum{a\left(\sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\right)}&\geq  \frac{2(k+1)a}{k}-\frac{1}{k}\left(\frac{a(b^2+1)}{ab+1}+\frac{a(c^2+1)}{ac+1}\right)\\
 &=\frac{6(k+1)}{k}-\frac{1}{k}\left[\sum{\frac{a(b^2+1)}{ab+1}}+\sum{\frac{a(c^2+1)}{ac+1}}\right]\\
 &=6
\end{align*}