Question

Cho a b lớn hơn 0 và a+b≤1tim min S=a/(1+b)+b/(1+a)+1/(a+b)

Answer 1

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

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

\(S\ge\frac{2\left(a+b\right)}{a+b+2}+\frac{1}{a+b}=2-\frac{4}{a+b+2}+\frac{1}{a+b}\)

Đặt \(a+b=t\Rightarrow0< t\le1\)

\(S\ge\frac{5}{3}+\frac{t+3}{3t}-\frac{4}{t+2}=\frac{5}{3}+\frac{t^2-7t+6}{3t\left(t+2\right)}=\frac{5}{3}+\frac{\left(6-t\right)\left(1-t\right)}{3t\left(t+2\right)}\ge\frac{5}{3}\)

\(S_{min}=\frac{5}{3}\) khi \(t=1\Leftrightarrow x=y=\frac{1}{2}\)