由均值不等式\[\sum_{cyc} a\sqrt{\frac{b}{a+2c}}\leqslant \sqrt{\frac{1}{3}\sum_{cyc}\frac{a^2b}{a+2c}}\]故\[\begin{split}&\sum_{cyc} a\sqrt{\frac{b}{a+2c}}\leqslant \frac{a+b+c}{\sqrt{3}}\\
\iff& \sum_{cyc}\frac{a^2b}{a+2c}\leqslant (\sum_{cyc}a)^2\\
\iff&\sum_{cyc}a^2+\sum_{cyc}ab\left(2-\frac{a}{a+2c} \right)=\sum_{cyc}a^2+\sum_{cyc}ab+2\sum_{cyc}\frac{abc}{a+2c}\geqslant0\end{split}\] |