Cho a, b, c>0 thỏa mãn: $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=6$
CMR: $\sum \frac{1}{3a+3b+2c}\leq \frac{3}{2}$
Đặt a+b=x;b+c=y;c+a=z ta có : $\sum \frac{1}{x}=6$
Mà : VT=$\sum \frac{1}{2x+y+z}\leq \frac{1}{16}(\sum \frac{4}{x})=\frac{1}{4}(\sum \frac{1}{x})=\frac{3}{2}$ ( đpcm )