ta có \(x\sqrt{a+y}+y\sqrt{a+x}=\sqrt{x}\sqrt{ax+xy}+\sqrt{y}\sqrt{ay+xy}\)
<=\(\sqrt{\left(x+y\right)\left(ax+xy+ay+xy\right)}=\sqrt{b\left[a\left(x+y\right)+2xy\right]}=\sqrt{b.a.b+b2xy}\)
Mà \(2xy\le\frac{\left(x+y\right)^2}{2}=\frac{b}{2}\Rightarrow b.2xy\le\frac{b^2}{2}\)
=>...\(\le\sqrt{b^2a+\frac{b^2}{2}}=b\sqrt{a+\frac{1}{2}}\)
Dâu = xảy ra <=> x=y=b/2
^_^