\(a,b>0;a+b=2\)
\(P=\dfrac{a}{\sqrt{4-a^2}}+\dfrac{b}{\sqrt{4-b^2}}=\dfrac{a^2}{a\sqrt{4-a^2}}+\dfrac{b^2}{b\sqrt{4-b^2}}\ge\dfrac{\left(a+b\right)^2}{a\sqrt{4-a^2}+b\sqrt{4-b^2}}\left(1\right)\)
Ta có: \(a\sqrt{4-a^2}+b\sqrt{4-b^2}\)
\(=a\sqrt{\left(2-a\right)\left(2+a\right)}+b\sqrt{\left(2-b\right)\left(2+b\right)}\)
\(=a\sqrt{b\left(2a+b\right)}+b\sqrt{a\left(2b+a\right)}\)
\(=\dfrac{1}{\sqrt{3}}\left[a\sqrt{3b\left(2a+b\right)}+b\sqrt{3a\left(2b+a\right)}\right]\)
\(\le\dfrac{1}{\sqrt{3}}.\left(a.\dfrac{3b+2a+b}{2}+b.\dfrac{3a+2b+a}{2}\right)\)
\(=\dfrac{1}{\sqrt{3}}.\left[a\left(a+2b\right)+b\left(b+2a\right)\right]\)
\(=\dfrac{1}{\sqrt{3}}\left(a^2+b^2+2ab+2ab\right)\)
\(=\dfrac{1}{\sqrt{3}}\left[\left(a+b\right)^2+2ab\right]\)
\(\le\dfrac{1}{\sqrt{3}}\left[\left(a+b\right)^2+2.\left(\dfrac{a+b}{2}\right)^2\right]=\dfrac{\sqrt{3}}{2}\left(a+b\right)^2\)
Do đó \(a\sqrt{4-a^2}+b\sqrt{4-b^2}\le\dfrac{\sqrt{3}}{2}\left(a+b\right)^2\left(2\right)\)
Từ (1), (2) suy ra: \(P\ge\dfrac{\left(a+b\right)^2}{\dfrac{\sqrt{3}}{2}\left(a+b\right)^2}=\dfrac{2\sqrt{3}}{3}\)
Dấu "=" xảy ra khi \(a=b=1\)
Vậy \(MinP=\dfrac{2\sqrt{3}}{3}\)
