Áp dụng BĐT AM - GM, ta có:
\(ab\le\dfrac{a^2+b^2}{2}=2\)
Áp dụng BĐT Cauchy Shwarz, ta có:
\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a+b\le\sqrt{2\left(a^2+b^2\right)}=2\sqrt{2}\)
Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\), ta có:
\(\dfrac{1}{M}=\dfrac{a+b+2}{ab}=\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{2}{ab}\)
\(\ge\dfrac{4}{a+b}+\dfrac{2}{2}\ge\dfrac{4}{2\sqrt{2}}+1=1+\sqrt{2}\)
\(\Rightarrow M\le\dfrac{1}{1+\sqrt{2}}=-1+\sqrt{2}\)
Dấu "=" xảy ra khi \(a=b=\sqrt{2}\)
Cách khác bạn nhé
\(M=\dfrac{ab}{a+b+2}=\dfrac{2ab}{2\left(a+b+2\right)}=\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2\left(a+b+2\right)}\)
\(=\dfrac{\left(a+b\right)^2-4}{2\left(a+b+2\right)}=\dfrac{a+b-2}{2}=\dfrac{\sqrt{\left(a+b\right)^2}-2}{2}\)
\(\le\dfrac{\sqrt{\left(a+b\right)^2+\left(a-b\right)^2}-2}{2}=\dfrac{\sqrt{2\left(a^2+b^2\right)-2}}{2}\)
\(=\dfrac{\sqrt{8}-2}{2}=\sqrt{2}-1\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=\sqrt{2}\)
