Có \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)(bđt cosi vs hai số dương)
=> 4\(\ge2ab\) <=> 2\(\ge ab\) <=> \(\frac{2}{ab}\ge1\) (*) => \(\frac{2}{\sqrt{ab}}\ge\sqrt{2}\)
AD bđt cosi vs hai số dương có:
\(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\) \(\ge\sqrt{2}\) (**).
Từ (*),(**) => \(\frac{1}{a}+\frac{1}{b}+\frac{2}{ab}\ge\sqrt{2}+1\)
Có \(M=\frac{ab}{a+b+2}=\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{2}{ab}}\le\frac{1}{\sqrt{2}+1}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a^2=b^2\\\frac{1}{a}=\frac{1}{b}\end{matrix}\right.< =>\left\{{}\begin{matrix}a=b\\a=b\end{matrix}\right.< =>a=b=\sqrt{2}\)(vì a,b>0)
Vậy maxM=\(\sqrt{2}-1\)
