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AD Cauchy-Schwarz:
$\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b} \ge \dfrac{(a+b+c)^2}{2(a+b+c)}=\dfrac{a+b+c}{2}$



Cách khác ;

$\dfrac{a^2}{b+c}+\dfrac{b+c}{4} \ge 2.\dfrac{a}{2}$

$\dfrac{b^2}{a+c}+\dfrac{a+c}{4} \ge 2.\dfrac{b}{2}$

$\dfrac{c^2}{a+b}+\dfrac{a+b}{4} \ge 2.\dfrac{c}{2}$

Cộng theo vế suy ra đpcm
 
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