Xét $P=\frac{25a^2}{a(b+c)}+\frac{4b^2}{b(a+c)}+\frac{9c^2}{c(a+b)}$Áp dụng schwarz $P\ge \frac{(5a+2b+3c)^2}{2(ab+ac+bc)}$
Ta cm $(5a+2b+3c)^2>24(ab+ac+bc)$
$\Leftrightarrow 25a^2+4b^2+9c^2-4ab+6ac-12bc>0$
$\Leftrightarrow 24a^2+(a+3c-2b)^2>0$ luôn đúng.
Ta có đpcm