có $\frac{bc}{a^{3}(c+2b)} + \frac{c+2b}{9abc}\geq \frac{2}{3a^{2}}$
$\frac{ca}{b^{3}(a+2c)} + \frac{a+2c}{9abc}\geq \frac{2}{3b^{2}}$
$\frac{ab}{c^{3}(b+2a)} + \frac{b+2a}{9abc} \geq \frac{2}{3c^{2}}$
cộng 3 bẩt đẳng thức, có $\sum \frac{bc}{a^{3}(c+2b)} + \sum \frac{c+2b}{9abc} \geq \sum \frac{2}{3a^{2}}$$\sum \frac{bc}{a^{3}(c+2b)} + \sum \frac{c+2b}{9abc} \geq \sum \frac{2}{3a^{2}} \geq \frac{2}{3} (\frac{1}{ab}+\frac{1}{bc}+ \frac{1}{ca})$
tương đường với $\sum \frac{bc}{a^{3}(c+2b)} \geq \frac{1}{3}(\sum \frac{1}{ab})$ = 1
- nolunne yêu thích