Question

Cho a, b là các số thực dương thỏa mãn \(a+b\le1\). Tìm GTNN của \(A=\dfrac{a}{b+1}+\dfrac{b}{a+1}+\dfrac{1}{a+b}\)

Answer 1

\(A=\dfrac{a}{b+1}+\dfrac{b}{a+1}+\dfrac{1}{a+b}\)

\(\ge\dfrac{a}{a+2b}+\dfrac{b}{2a+b}+\dfrac{1}{a+b}\)

\(=\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{2ab+b^2}+\dfrac{1}{a+b}\)

\(\ge\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+2ab}+\dfrac{1}{a+b}\)

\(\ge\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+\dfrac{\left(a+b\right)^2}{2}}+\dfrac{1}{a+b}\)

\(=\dfrac{\left(a+b\right)^2}{\dfrac{3}{2}\left(a+b\right)^2}+\dfrac{1}{a+b}=\dfrac{2}{3}+\dfrac{1}{a+b}\ge\dfrac{2}{3}+1=\dfrac{5}{3}\)

\("="\Leftrightarrow a=b=\dfrac{1}{2}\)