\(\frac{ab}{a+b+2}=\frac{1}{\frac{1}{b}+\frac{1}{a}+\frac{2}{ab}}\le\frac{1}{\frac{4}{a+b}+\frac{2}{ab}}\le\frac{1}{\frac{4}{\sqrt{2\left(a^2+b^2\right)}}+\frac{2}{\frac{a^2+b^2}{2}}}\le\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
dấu bằng xảy ra khi a=b=\(\sqrt{2}\)