Question

Cho a,b >0 thỏa mãn \(a+b\ge2.\)Tìm GTLN của \(M=\dfrac{1}{a+b^2}+\dfrac{1}{b+a^2}\)

Answer 1

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

Tương tự: \(\dfrac{1}{b+a^2}\le\dfrac{b+1}{\left(a+b\right)^2}\)

\(\Rightarrow M\le\dfrac{a+b+2}{\left(a+b\right)^2}=\dfrac{2}{\left(a+b\right)^2}+\dfrac{1}{a+b}=\dfrac{2}{\left(a+b\right)^2}+\dfrac{1}{a+b}-1+1\)

\(\Rightarrow M\le\left(\dfrac{2}{a+b}-1\right)\left(\dfrac{1}{a+b}+1\right)+1=\left(\dfrac{2-a-b}{a+b}\right)\left(\dfrac{1}{a+b}+1\right)+1\le1\)

\(M_{max}=1\) khi \(a=b=1\)