Giải thích các bước giải:
Ta có :
$a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{(a^2+b^2)(b+1+a+1)}$ BĐT bunhia
$\to a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{(a+b+2}, a^2+b^2=1$
$\to a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{\sqrt{(a+b)^2}+2}$
$\to a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{\sqrt{2(a^2+b^2)}+2}$
$\to a\sqrt{b+1}+b\sqrt{a+1}\le\sqrt{\sqrt{2}+2}$
Dấu = xảy ra khi $a=b=\dfrac{1}{\sqrt{2}}$