Giải
Ta có:
$\sqrt{\dfrac{2a}{a + b}} + \sqrt{\dfrac{2b}{b + c}} + \sqrt{\dfrac{2c}{c + a}}$
$= \dfrac{a^2}{\sqrt{\dfrac{a^3(a + b)}{2}}} + \dfrac{b^2}{\sqrt{\dfrac{b^3(b + c)}{2}}} + \dfrac{c^2}{\sqrt{\dfrac{c^3(c + a)}{2}}} \geq \dfrac{(a + b + c)^2}{\sqrt{\dfrac{a^3(a + b)}{2}} + \sqrt{\dfrac{b^3(b + c)}{2}} + \sqrt{\dfrac{c^3(c + a)}{2}}}$
Ta sẽ chứng minh:
$\sqrt{\dfrac{a^3(a + b)}{2}} + \sqrt{\dfrac{b^3(b + c)}{2}} + \sqrt{\dfrac{c^3(c + a)}{2}} \leq a^2 + b^2 + c^2 \, (1)$
Thật vậy:
$\sqrt{\dfrac{a^3(a + b)}{2}} = \sqrt{\dfrac{a^2(a^2 + ab)}{2}} \leq \dfrac{3a^2 + ab}{4}$
Từ đó suy ra:
$VT_{(1)} \leq \dfrac{3(a^2 + b^2 + c^2) + ab + ac + bc}{4} \leq a^2 + b^2 + c^2$
Do đó, ta có điều phải chứng minh.