11. (15 points) Given $a, b>0$. Find the maximum values of $\sqrt{\frac{a}{2 a+b}}$ $+\sqrt{\frac{b}{2 b+a}}$ and $\sqrt{\frac{a}{a+2 b}}+\sqrt{\frac{b}{b+2 a}}$.
11. Solution: By Cauchy-Schwarz inequality, we have
$$
\sqrt{\frac{a}{2 a+b}}+\sqrt{\frac{b}{2 b+a}} \leqslant
$$
$$
\sqrt{2\left(\frac{a}{2 a+b}+\frac{b}{2 b+a}\right)}
$$

Without loss of generality, let \( a + b = 1 \), then the right side of (1) is
$$
\begin{array}{c}
\sqrt{2\left(\frac{a}{1+a}+\frac{b}{1+b}\right)}=\sqrt{4-2\left(\frac{1}{1+a}+\frac{1}{1+b}\right)}, \\
\because\left(\frac{1}{1+a}+\frac{1}{1+b}\right)(1+a+1+b) \geqslant 4, \\
\therefore \frac{1}{1+a}+\frac{1}{1+b} \geqslant \frac{4}{3},
\end{array}
$$
\(\therefore\) the right side of (1) \(\leqslant \sqrt{4-\frac{8}{3}}=\frac{2}{3} \sqrt{3}\).
Similarly, \(\sqrt{\frac{a}{a+2 b}}+\sqrt{\frac{b}{b+2 a}}\) can be solved.