[Proof] Represent the polynomial $P(x)$ as
$$P(x)=a F_{1}(x) \cdots F_{m}(x) G_{1}(x) \cdots G_{k}(x)$$

where $F_{i}(x)$ and $G_{j}(x)$ are
$$\begin{array}{l} F_{i}(x)=x-\alpha_{i} ; \alpha_{i} \leqslant 0, i=1,2, \cdots, m \\ G_{j}(x)=\left(x-\beta_{j}\right)\left(x-\overline{\beta_{j}}\right), \beta_{j} \in C \backslash R, j=1,2, \cdots, k \end{array}$$

and $a>0$. Note that the product of polynomials with non-negative coefficients is still a polynomial with non-negative coefficients. Therefore, it suffices to prove that each polynomial $F_{i}(x)$ and $G_{j}(x)$ can be expressed as the quotient $\frac{Q(x)}{R(x)}$ of polynomials with non-negative coefficients. For the polynomial $F(x)=x-\alpha, \alpha \leqslant 0$, take
$$Q(x)=F(x)=x+|\alpha|, R(x)=1$$

Next, if $\operatorname{Re} \beta \leqslant 0$, then for the polynomial
$$G(x)=(x-\beta)(x-\bar{\beta}),$$

let $\quad Q(x)=G(x)=x^{2}+2|\operatorname{Re} \beta| x+|\beta|^{2}, R(x)=1$;
if $\operatorname{Re} \beta>0$, then assume $\arg \beta \in\left(0, \frac{\pi}{2}\right)$, take $n \in N$, such that
$$2^{n} \arg \beta \in\left(\frac{\pi}{2}, \pi\right)$$

and when $S=0,1, \cdots, n-1$,
$$2^{S} \arg \beta \in\left(0, \frac{\pi}{2}\right)$$

Since $\quad \arg \left(\beta^{2^{r}}\right)=2^{r} \arg \beta \quad(0 \leqslant r \leqslant n)$,
it follows that $\operatorname{Re}\left(\beta^{2^{2}}\right) \leqslant 0$; and when $S=0,1, \cdots, n-1$, $\operatorname{Re}\left(\beta^{2^{*}}\right)>0$, thus we have
$$\begin{aligned} G(x) & \equiv(x-\beta)(x-\bar{\beta}) \\ & \equiv \frac{\left(x^{2}-\beta^{2}\right)\left(x^{2}-\bar{\beta}^{2}\right)}{(x+\beta)(x+\bar{\beta})} \\ & \equiv \frac{\left(x^{4}-\beta^{4}\right)\left(x^{4}-\bar{\beta}^{4}\right)}{(x+\beta)(x+\bar{\beta})\left(x^{2}+\beta^{2}\right)\left(x^{2}+\bar{\beta}^{2}\right)} \\ & \equiv \cdots \\ & \equiv \frac{\left(x^{2^{n}}-\beta^{2^{n}}\right)\left(x^{2^{n}}-\bar{\beta}^{2^{n}}\right)}{(x+\beta)(x+\bar{\beta}) \cdots\left(x^{2^{n-1}}+\beta^{2^{n-1}}\right)\left(x^{2^{n-1}}+\bar{\beta}^{2^{n-1}}\right)} \\ & \equiv \frac{Q(x)}{R(x)} \end{aligned}$$

where
$$\begin{array}{l} Q(x)=\left(x^{2^{n}}-\beta^{2^{n}}\right)\left(x^{2^{n}}-\bar{\beta}^{2^{n}}\right) \\ R(x)=(x+\beta)(x+\bar{\beta}) \cdots\left(x^{2^{n-1}}+\beta^{2^{n-1}}\right)\left(x^{2^{n-1}}+\bar{\beta}^{2^{n-1}}\right) \end{array}$$

are polynomials with non-negative coefficients, because all polynomials
$$\begin{aligned} & \left(x^{l}+\gamma\right)\left(x^{l}+\bar{\gamma}\right) \\ \equiv & x^{2 l}+(2 \operatorname{Re} \gamma) x^{l}+|r|^{2}, l \in N, \operatorname{Re} r>0 \end{aligned}$$

have non-negative coefficients.