Câu hỏi 75
Cho $a$, $b$, $c \in (0, \infty)$ thỏa mãn $\sqrt{a} + \sqrt{b} + \sqrt{c} \le 2$. Chứng minh rằng
$$
\frac{\sqrt{a^2 - ab + b^2}}{a^2 + b^2} + \frac{\sqrt{b^2 - bc + c^2}}{b^2 + c^2} + \frac{\sqrt{c^2 - ca + a^2}}{c^2 + a^2} \le \frac{1}{\sqrt{abc}}
$$
$$
\frac{\sqrt{a^2 - ab + b^2}}{a^2 + b^2} + \frac{\sqrt{b^2 - bc + c^2}}{b^2 + c^2} + \frac{\sqrt{c^2 - ca + a^2}}{c^2 + a^2} \le \frac{1}{\sqrt{abc}}
$$