设 $0<a<b$, 且 $a+b=1$, 在下列四个数中最大的是 ( )
解析 $\because a b<\left(\frac{a+b}{2}\right)^{2}, \therefore a b<\frac{1}{4}, \therefore 2 a b<\frac{1}{2}$.

$\because \sqrt{\frac{a^{2}+b^{2}}{2}}>\frac{a+b}{2}>0, \therefore \sqrt{\frac{a^{2}+b^{2}}{2}}>\frac{1}{2}$,

$\therefore a^{2}+b^{2}>\frac{1}{2}$.

$\because b-\left(a^{2}+b^{2}\right)=\left(b-b^{2}\right)-a^{2}=b(1-b)-a^{2}$

$=a b-a^{2}=a(b-a)>0, \quad \therefore b>a^{2}+b^{2}, \quad \therefore b$ 最大.