1. Given point $P(1,2)$ is inside (including the boundary) the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and outside (including the boundary) the circle $x^{2}+y^{2}=\frac{a^{2}+2 b^{2}}{3}$. If $a, b \in \mathbf{R}_{+}$. Then the minimum value of $a+b$ is ( ).
(A) $\sqrt{5}+2$
(B) $\sqrt{6}+\sqrt{3}$
(C) $2 \sqrt{5}$
(D) $\sqrt{3}+\sqrt{5}$
- 1.B.

From the given, we have $\frac{1}{a^{2}}+\frac{4}{b^{2}} \leqslant 1$ and $a^{2}+2 b^{2} \leqslant 15$. Therefore,
$$
\frac{b^{2}}{b^{2}-4} \leqslant a^{2} \leqslant 15-2 b^{2} \text {. }
$$

Simplifying, we get $\left(b^{2}-5\right)\left(b^{2}-6\right) \leqslant 0$.
Thus, $5 \leqslant b^{2} \leqslant 6$.
Hence, $a+b \geqslant \sqrt{\frac{b^{2}}{b^{2}-4}}+b$
$$
\begin{array}{l}
=\sqrt{\frac{t+4}{t}}+\sqrt{t+4}\left(t=b^{2}-4 \in[1,2]\right) . \\
\text { and } \sqrt{\frac{t+4}{t}}+\sqrt{t+4} \geqslant \sqrt{3}+\sqrt{6} \\
\Leftrightarrow \sqrt{t+4}-\sqrt{6} \geqslant \sqrt{3}-\sqrt{\frac{t+4}{t}} \\
\Leftrightarrow \frac{t-2}{\sqrt{t+4}+\sqrt{6}} \geqslant \frac{2(t-2)}{\sqrt{3} t+\sqrt{t(t+4)}} .
\end{array}
$$

Since $t \leqslant 2$, it is equivalent to
$$
2(\sqrt{t+4}+\sqrt{6}) \geqslant \sqrt{3} t+\sqrt{t(t+4)} \text {. }
$$

Given $2 \sqrt{6} \geqslant \sqrt{3} t$ and $2 \geqslant \sqrt{t}$, we know the above inequality holds.
Therefore, $a+b \geqslant \sqrt{3}+\sqrt{6}$.
When $a^{2}=3, b^{2}=6$, the conditions are met and the minimum value is achieved.