11. If the inequality $\frac{1}{\sqrt{20 a+23 b}}+\frac{1}{\sqrt{23 a+20 b}} \geqslant \frac{\lambda}{\sqrt{a+b}}$ holds for all positive real numbers $a, b$, find the maximum value of $\lambda$.
Solve:
$\frac{1}{\sqrt{20 a+23 b}}+\frac{1}{\sqrt{23 a+20 b}} \geqslant \frac{4}{\sqrt{20 a+23 b}+\sqrt{23 a+20 b}}$
$\geqslant \frac{4}{\sqrt{2(20 a+23 b+23 a+20 b)}}=\frac{4}{\sqrt{86} \cdot \sqrt{a+b}}$, equality holds when $a=b$.
Therefore, the maximum value of $\lambda$ is $\frac{4}{\sqrt{86}}=\frac{2 \sqrt{86}}{43}$.