Since $\Delta = (\sqrt{m})^2 - 4(m+1) < 0$, we have $f(x) > 0$.
We use mathematical induction to prove that the statement holds when $n = 2^k$.

a. When $k=1$, $n=2$. Since
$$\sqrt{m}x_1x_2(\sqrt{x_1}-\sqrt{x_2})^2 + (m+1)(x_1-x_2)^2 + \sqrt{m}(m+1)(\sqrt{x_1}-\sqrt{x_2})^2 \ge 0,$$
we have
$$(x_1x_2 + \sqrt{mx_1x_2} + m + 1)^2 \le (x_1^2 + \sqrt{m}x_1 + m + 1)(x_2^2 + \sqrt{m}x_2 + m + 1),$$
that is, $f(\sqrt{x_1x_2}) \le \sqrt{f(x_1)f(x_2)}$, and the necessary and sufficient condition for equality to hold is $x_1 = x_2$.

b. Suppose the statement holds for $n = 2^k$. Then for $n = 2^{k+1}$,
$$\begin{align*} & f(\sqrt[2^{k+1}]{x_1x_2\cdots x_{2^{k+1}}}) \\ &= f(\sqrt[2^k]{x_1x_2\cdots x_{2^k}} \sqrt[2^k]{x_{2^k+1}x_{2^k+2}\cdots x_{2^{k+1}}}) \\ &\le \sqrt{f(\sqrt[2^k]{x_1x_2\cdots x_{2^k}}) f(\sqrt[2^k]{x_{2^k+1}x_{2^k+2}\cdots x_{2^{k+1}}})} \\ &= \sqrt[2^{k+1}]{f(x_1)f(x_2)\cdots f(x_{2^{k+1}})}. \end{align*}$$

For any positive integer $n$, there must exist $k$ such that $2^k \le n < 2^{k+1}$. Let $G = \sqrt[n]{x_1x_2\cdots x_n}$,
then
$$f(G) = f(\sqrt[2^{k+1}]{x_1 x_2 \cdots x_n\, G \cdots G}) \le \sqrt[2^{k+1}]{f(x_1)f(x_2)\cdots f(x_n)[f(G)]^{2^{k+1}-n}},$$
that is, $[f(G)]^n \le f(x_1)f(x_2)\cdots f(x_n)$,
$$f(\sqrt[n]{x_1x_2\cdots x_n}) \le \sqrt[n]{f(x_1)f(x_2)\cdots f(x_n)}.$$
The necessary and sufficient condition for equality to hold is $x_1 = x_2 = \cdots = x_n$.