Ta có:
\(a^2+b^2=4\Leftrightarrow a^2+2ab+b^2=4+2ab\Leftrightarrow\left(a+b\right)^2-4=2ab\)
\(\Leftrightarrow\left(a+b-2\right)\left(a+b+2\right)=2ab\)
\(\Leftrightarrow\dfrac{\left(a+b-2\right)\left(a+b+2\right)}{2}=ab\)
\(\Rightarrow M=\dfrac{ab}{a+b+2}=\dfrac{\left(a+b-2\right)\left(a+b+2\right)}{2\left(a+b+2\right)}=\dfrac{a+b-2}{2}=\dfrac{a}{2}+\dfrac{b}{2}-\dfrac{2}{2}=\dfrac{a}{2}+\dfrac{b}{2}-1\)
A/dụng bđt bunhiacopxki có:
\(\left(\dfrac{a}{2}+\dfrac{b}{2}\right)^2\le\left[\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2\right]\left(a^2+b^2\right)=\dfrac{1}{2}\cdot4=2\)
\(\Rightarrow\dfrac{a}{2}+\dfrac{b}{2}\le\sqrt{2}\Rightarrow\dfrac{a}{2}+\dfrac{b}{2}-1\le\sqrt{2}-1\)
hay \(M\le\sqrt{2}-1\)
Dấu ''='' xảy ra khi \(a=b=\sqrt{2}\)
Vậy \(Max_M=\sqrt{2}-1\)<=> a=b=\(\sqrt{2}\)
\(a^2+b^2=4\Leftrightarrow\left(a+b\right)^2=4+2ab\Leftrightarrow\left(a+b\right)^2-4=2ab\)
=> \(2M=\dfrac{2ab}{a+b+2}=\dfrac{\left(a+b\right)^2-4}{a+b+2}=\dfrac{\left(a+b+2\right)\left(a+b-2\right)}{a+b+2}=a+b-2\)
Ta có: \(a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2\cdot4}=2\sqrt{2}\left(bunhiacopxki\right)\)
\(\Rightarrow2M\le2\sqrt{2}-2\Rightarrow M\le\sqrt{2}-1\)
đẳng thức xảy ra <=> a=b=√2
vậy...