CMR:
$\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}\geq \frac{51}{7}$
giải như sau:
$\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}\geq \frac{51}{7}\Leftrightarrow \frac{2y+3z+5}{1+x}+1+\frac{3z+x+5}{1+2y}+1+\frac{x+2y+5}{1+3z}+1\geq \frac{51}{7}+3=\frac{72}{7} \Leftrightarrow \frac{2y+3z+6+x}{1+x}+\frac{3z+x+2y+6}{1+2y}+\frac{x+2y+3z+6}{1+3z}= (3z+x+2y+6)(\frac{1}{1+x}+\frac{1}{1+2y}+\frac{1}{1+3z})\geqslant (18+6)(\frac{9}{3+x+2y+3z})=24.\frac{3}{7}=\frac{72}{7}$