11. (20 points) Given positive real numbers $a, b$ satisfy $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1, a^{2}+2 b^{2} \leqslant 15$.
Find the range of $a+b$.
11. From $a^{2}+2 b^{2} \leqslant 15 \Rightarrow a^{2} \leqslant 15-2 b^{2}$.

From $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1 \Rightarrow \frac{1}{15-2 b^{2}}+\frac{4}{b^{2}} \leqslant 1$
$\Rightarrow b^{4}-11 b^{2}+30 \leqslant 0 \Rightarrow 5 \leqslant b^{2} \leqslant 6$.
Also, $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1 \Rightarrow a \geqslant \frac{b}{\sqrt{b^{2}-4}}$.
Thus, $a+b \geqslant \frac{b}{\sqrt{b^{2}-4}}+b$.
Considering the function $f(x)=\frac{x}{\sqrt{x^{2}-4}}+x$ on the interval $[\sqrt{5}, \sqrt{6}]$,
we know
$$
f^{\prime}(x)=1-\frac{4}{\sqrt{\left(x^{2}-4\right)^{3}}}<0 .
$$

Therefore, $a+b \geqslant \frac{b}{\sqrt{b^{2}-4}}+b \geqslant \sqrt{3}+\sqrt{6}$.
When $b=\sqrt{6}, a=\sqrt{3}$, the equality holds.
From $a \leqslant \sqrt{15-2 b^{2}} \Rightarrow a+b \leqslant \sqrt{15-2 b^{2}}+b$.
Considering the function $g(x)=\sqrt{15-2 x^{2}}+x$ on the interval $[\sqrt{5}, \sqrt{6}]$,
we know
$$
g^{\prime}(x)=1-\frac{x}{\sqrt{15-2 x^{2}}} \leqslant 0 .
$$

Therefore, $a+b \leqslant \sqrt{15-2 b^{2}}+b \leqslant 2 \sqrt{5}$.
When $b=a=\sqrt{5}$, the equality holds.
In summary, the range of $a+b$ is $[\sqrt{3}+\sqrt{6}, 2 \sqrt{5}]$.