Question

Answer 1

Lời giải:
Áp dụng BĐT Cô-si:

$Q=a^2+\frac{2}{3a}=\frac{80a^2}{81}+\frac{a^2}{81}+\frac{1}{3a}+\frac{1}{3a}$

$\geq \frac{80a^2}{81}+3\sqrt[3]{\frac{a^2}{81}.\frac{1}{3a}.\frac{1}{3a}}$

$\geq \frac{80.3^2}{81}+\frac{1}{3}=\frac{83}{9}$

Vậy $Q_{\min}=\frac{83}{9}$ khi $a=3$

Answer 2

Lời giải:
Áp dụng BĐT Cô-si:

$Q=a^2+\frac{2}{3a}=\frac{80a^2}{81}+\frac{a^2}{81}+\frac{1}{3a}+\frac{1}{3a}$

$\geq \frac{80a^2}{81}+3\sqrt[3]{\frac{a^2}{81}.\frac{1}{3a}.\frac{1}{3a}}$

$\geq \frac{80.3^2}{81}+\frac{1}{3}=\frac{83}{9}$

Vậy $Q_{\min}=\frac{83}{9}$ khi $a=3$