$\dfrac{a^2}{m} + \dfrac{b^2}{n} \geq \dfrac{(a+b)^2}{m+n}$
$\Leftrightarrow \dfrac{a^2n + b^2m}{mn} \geq \dfrac{a^2 + b^2 + 2ab}{m+ n}$
$\Leftrightarrow (a^2n + b^2m)(m+n) \geq (a^2 +b^2 +2ab)mn$
$\Leftrightarrow a^2mn + a^2n^2 + b^2m^2 + b^2mn \geq a^2mn + b^2mn + 2abmn$
$\Leftrightarrow a^2n^2 - 2abmn + b^2m^2 \geq 0$
$\Leftrightarrow (an - bm)^2 \geq 0$ (luôn đúng)
Vậy $\dfrac{a^2}{m} + \dfrac{b^2}{n} \geq \dfrac{(a+b)^2}{m+n}$
