10. Find the maximum value of the positive real number $k$ such that for any positive real numbers $a, b$, we have
$$
\sqrt{a^{2}+k b^{2}}+\sqrt{b^{2}+k a^{2}} \geq a+b+(k-1) \sqrt{a b}
$$
10. Analysis: $k=3$. When $a=b$, $2 \sqrt{k+1} a \geq(k+1) a$ holds, i.e., $k \leq 3$;

Next, we prove that for any positive real numbers $a, b$, the inequality $\sqrt{a^{2}+3 b^{2}}+\sqrt{b^{2}+3 a^{2}} \geq a+b+2 \sqrt{a b}$ holds.
Introduce the parameter $p$. By the Cauchy-Schwarz inequality, we have:
$\sqrt{\left(a^{2}+3 b^{2}\right)\left(1+p^{2}\right)} \geq a+\sqrt{3} p b$, i.e., $\sqrt{\left(a^{2}+3 b^{2}\right)} \geq \frac{a+\sqrt{3} p b}{\sqrt{1+p^{2}}}$,
Similarly, $\sqrt{\left(b^{2}+3 a^{2}\right)} \geq \frac{b+\sqrt{3} p a}{\sqrt{1+p^{2}}}$;
Adding the two inequalities, we get: $\sqrt{a^{2}+3 b^{2}}+\sqrt{b^{2}+3 a^{2}} \geq \frac{(\sqrt{3} p+1)}{\sqrt{1+p^{2}}}(a+b)$,
Since $2(a+b) \geq(\sqrt{a}+\sqrt{b})^{2}=a+b+2 \sqrt{a b}$, it suffices to show that $\frac{\sqrt{3} p+1}{\sqrt{1+p^{2}}} \geq 2$ has a solution.
It is easy to see that when $p=\sqrt{3}$, $\frac{\sqrt{3} p+1}{\sqrt{1+p^{2}}}=2$ holds, i.e., $\sqrt{a^{2}+3 b^{2}}+\sqrt{b^{2}+3 a^{2}} \geq 2(a+b) \geq a+b+2 \sqrt{a b}$.