10. (20 points) Given positive numbers $a, b, c$ satisfying $a b + b c + c a + 2 a b c = 1$.

Find the maximum value of $\sqrt{a b} + \sqrt{b c} + \sqrt{c a}$.
10. Let $x=\sqrt{a b}, y=\sqrt{b c}, z=\sqrt{c a}$. Then
$$
x^{2}+y^{2}+z^{2}+2 x y z=1 \text {. }
$$

Let $x+y+z=t$, and combine with equation (1) to eliminate $x$, we get
$$
2 y^{2}(1-z)+2(z-t)(1-z) y+(t-z)^{2}+z^{2}-1=0 \text {. }
$$

Given $0<z<1$, the necessary and sufficient condition for the equation in $y$ to have a solution is
$$
\begin{aligned}
\Delta_{y}= & 4(z-t)^{2}(1-z)^{2}- \\
& 8(1-z)\left[(t-z)^{2}+z^{2}-1\right] \geqslant 0 \\
\Rightarrow & z^{2}+2(1-t) z+t^{2}-2 \leqslant 0 \\
\Rightarrow & \Delta_{z}=4(1-t)^{2}-4\left(t^{2}-2\right) \\
= & -8 t+12 \geqslant 0 \\
\Rightarrow & t \leqslant \frac{3}{2} .
\end{aligned}
$$

The equality holds if and only if $x=y=z=\frac{1}{2}$ or $a=b=c=\frac{1}{2}$. Therefore, the maximum value of $t$ is $\frac{3}{2}$.