Áp dụng BĐT cosi cho 2 số dương
\(1=a^2+b^2\ge2ab\Leftrightarrow ab\le\dfrac{1}{2}\)
Mà \(\left(a+b\right)^2=1+2ab\le1+2\cdot\dfrac{1}{2}=2\Leftrightarrow a+b\le\sqrt{2}\)
Áp dụng BĐT Bunhiacopski
\(\left(a\sqrt{1+b}+b\sqrt{1+a}\right)^2\le\left(a^2+b^2\right)\left(1+b+1+a\right)=2+a+b\le2+\sqrt{2}\\ \Leftrightarrow a\sqrt{1+b}+b\sqrt{1+a}\le\sqrt{2+\sqrt{2}}\)
Dấu \("="\Leftrightarrow\dfrac{a}{b}=\sqrt{\dfrac{1+b}{1+a}}\Leftrightarrow a=b=\dfrac{1}{2}\)
Áp dụng BĐT Bunhicopski:
\(\left(a\sqrt{1+b}+b\sqrt{1+a}\right)\le\left(a^2+b^2\right)\left(1+b+1+a\right)=a+b+2\left(1\right)\)
Ta có: \(a^2+b^2\ge2ab\)(BĐT Cauchy)
\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow\left(a+b\right)^2\le2\Rightarrow a+b\le\sqrt{2}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left(a\sqrt{1+b}+b\sqrt{1+a}\right)^2\le2+\sqrt{2}\)
\(\Rightarrow a\sqrt{1+b}+b\sqrt{1+a}\le\sqrt{2+\sqrt{2}}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{\sqrt{2}}{2}\)
