设$a,b,c>0$,$a^2+b^2+c^2=3$证明
\[ \sqrt{\frac{a}{a^2+b^2+1}}+\sqrt{\frac{b}{b^2+c^2+1}}+\sqrt{\frac{c}{c^2+a^2+1}}\leq \sqrt{3}\]
证明
由Cauchy-Schwarz不等式
\[ \left(\sum\sqrt{\frac{a}{a^2+b^2+1}}\right)^{2}\leq \sum(a^2+c^2+1)\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}=9\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}\]
下面证明
\[\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}\leq \frac{1}{3} \]
\[ \Leftrightarrow 12\sum{a}-3\sum{a^3}\leq 34-a^2b^2c^2-2\sum{a^4}\]
由$a^2b^2c^2\leq 1$,只要证明
\[ 12\sum{a}-3\sum{a^3}\leq 33-2\sum{a^4}\]
\[ \Leftrightarrow \sum_{cyc}(1-a^2)\left(2a^3-3a+2+\frac{9}{a+1}\right)\geq 0 \]
记$ \sum(1-a^2)S_{a}$,易得$S_{c}\geq S_{b}\geq S_{a} $
由切比雪夫不等式
\[ \sum(1-a^2)S_{a}\geq \frac{1}{3}\sum(1-a^2)\sum S_{a}=0 \]
Done!

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