Cách khác nè:
$\frac{(a+1)^{6}}{b^{5}}+\frac{(b+1)^{6}}{a^{5}}\geq \frac{(2\sqrt{a})^{6}}{b^{5}}+\frac{(2\sqrt{b})^{6}}{a^{5}}=64(\frac{a^{3}}{b^{5}}+b+b)+64(\frac{b^{3}}{a^{5}}+a+a)-128(a+b)\geq 64.3.\frac{a}{b}+64.3.\frac{b}{a}-128.2\geq 64.3.2-128.2=128$
- quangtq1998, Master Kaiser và lily evans thích