Tìm gì hả em
a và b hay là tổng
\(VT=\frac{a}{a+2b}+\frac{b}{b+2a}+1\ge\frac{\left(a+b\right)^2}{a^2+b^2+4ab}+1\)
=> \(VT\ge\frac{\left(a+b\right)^2}{\left(a+b\right)^2+2ab}+1\ge\frac{\left(a+b\right)^2}{\left(a+b\right)^2+\frac{1}{2}\left(a+b\right)^2}+1=\frac{2}{3}+1=\frac{5}{3}\)
MinVT=5/3 khi a=b=1/2