9. (16 points) Find the maximum possible value of the real number $k$ such that the inequality $\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqslant \sqrt{a b}$ holds for any positive real numbers $a, b$.

Try to find the maximum possible value of the real number $k$ such that the inequality $\frac{2\left(a^{2}+k a b+b^{2}\right)}{(k+2)(a+b)} \geqslant \sqrt{a b}$ holds for any positive real numbers $a, b$.
9. The maximum possible value of $k$ is 6.

If $k=2$, the inequality is equivalent to $\frac{a+b}{2} \geqslant \sqrt{a b}$.
Now consider the case where $k>2$, then $k+2>0$, the original inequality is equivalent to
$$
\begin{array}{l}
2\left(a^{2}+k a b+b^{2}\right) \geqslant(k+2)(a+b) \sqrt{a b} \\
\Leftrightarrow 2\left(\frac{a^{2}}{b^{2}}+k \frac{a}{b}+1\right) \geqslant(k+2)\left(\frac{a}{b}+1\right) \sqrt{\frac{a}{b}} .
\end{array}
$$

Let $x=\sqrt{\frac{a}{b}}$, then the problem is equivalent to finding the largest $k$ such that
$$
2\left(x^{4}+k x^{2}+1\right) \geqslant(k+2)\left(x^{2}+1\right) x
$$

holds for any positive $x$.
We have
$$
\begin{array}{l}
k\left(\left(x^{2}+1\right) x-2 x^{2}\right) \leqslant 2\left(x^{4}+1-\left(x^{2}+1\right) x\right) \\
\Leftrightarrow k\left(x^{3}-2 x^{2}+x\right) \leqslant 2\left(x^{4}-x^{3}-x+1\right) \\
\Leftrightarrow k x(x-1)^{2} \leqslant 2(x-1)^{2}\left(x^{2}+x+1\right),
\end{array}
$$

with equality when $k=1$.
If $x \neq 1$, then
$$
k \leqslant \frac{2\left(x^{2}+x+1\right)}{x}=2+2\left(x+\frac{1}{x}\right),
$$

and the range of the right-hand side is $(6,+\infty)$, so $k \leqslant 6$.
In conclusion, the maximum value of $k$ is 6.