áp dụng bđt bunhiacopxki ta có
\frac{25a}{b+c}+25+\frac{16b}{a+c}+16+\frac{c}{a+b}+1=(a+b+c)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})
=\frac{1}{2}(b+c+a+c+a+b)(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b})
\geq \frac{(5+4+1)^{2}}{2}=50
\Rightarrow \frac{25a}{b+c}+ \frac{16b}{a+c}+ \frac{c}{a+b} \geq 8
ta thấy dấu "=" k xảy ra \Rightarrow đpcm