Question

Cho a,b là 2 số thực không âm thỏa mãn: \(a+b\le2\). Chứng minh:\(\dfrac{2+a}{1+a}+\dfrac{1-2b}{1+2b}\ge\dfrac{8}{7}\)

Answer 1

\(VT=1+\dfrac{1}{1+a}+\dfrac{2}{1+2b}-1=2\left(\dfrac{1}{2+2a}+\dfrac{1}{1+2b}\right)\)

\(VT\ge\dfrac{8}{3+2\left(a+b\right)}\ge\dfrac{8}{3+2.2}=\dfrac{8}{7}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=\dfrac{3}{4}\\b=\dfrac{5}{4}\end{matrix}\right.\)