Ta đặt: $x=b+c ;y=a+c ;z=a+b$ $\Rightarrow a=\frac{y+z-x}{2};b=\frac{x+z-y}{2};c=\frac{x+y-z}{2}$
Áp dụng BĐT Côsi, ta có:
$VT=\frac{25(y+z-x)}{2x}+\frac{16(x+z-y)}{2y}+\frac{x+y-z}{2z}=(\frac{25y}{2x}+\frac{8x}{y})+(\frac{25z}{2x}+\frac{x}{2z})+(\frac{8z}{y}+\frac{y}{2z})-21\geq 2\sqrt{\frac{25y}{2x}.\frac{8x}{y}}+2\sqrt{\frac{25z}{2x}.\frac{x}{2z}}+2\sqrt{\frac{8z}{y}.\frac{y}{2z}}-21=29-21=8$ (đpcm)