ta có :$\frac{2+a}{1+a}-1+\frac{1-2b}{1+2b}+1=\frac{2}{2+2a}+\frac{2}{2+2b}$rồi dùng $\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$ là xong
ta có :$\frac{2+a}{1+a}-1+\frac{1-2b}{1+2b}+1=\frac{2}{2+2a}+\frac{2}{2+2b}$roi dung $\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$ laf xong
ta có :$\frac{2+a}{1+a}-1+\frac{1-2b}{1+2b}+1=\frac{2}{2+2a}+\frac{2}{2+2b}$r
ồi d
ùng $\frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$ l
à xong