請教一不等式

kezhulu

想了幾天都沒什麼頭緒

gxqcn

\(\because \dfrac{2a^2+b^2-3c^2}{3c}=\dfrac{2a^2+b^2}{3c}-c,\qquad \dfrac{b(a-c)}{a+2b}+c=\dfrac{ab+bc+ca}{a+2b}\)

\(\therefore\) 原式等价于：\[2\left(\frac{a^2}{c}+\frac{c^2}{b}+\frac{b^2}{a}\right)+\left(\frac{b^2}{c}+\frac{a^2}{b}+\frac{c^2}{a}\right) \geq 3\left(ab+bc+ca\right)\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\]