Theo BĐT Bunyakovsky ta có:
[tex](1^2+1^2)(a^2+b^2)\geq (a+b)^2\Rightarrow a+b\leq \sqrt{2(a^2+b^2)}=2\sqrt{2}[/tex]
[tex]2M=\frac{2ab}{a+b+2}=\frac{(a+b)^2-4}{a+b+2}=a+b-2\leq 2\sqrt{2}-2\\\Rightarrow M\leq \sqrt{2}-1[/tex]
Dấu = xảy ra khi [tex]a=b=\sqrt{2}[/tex]