Assume that $a \ge b \ge c$. Set $G(x) = P(x)/x^n$ and notice that $G(a) \le G(b) \le G(c)$.

Let, without loss of generality, $a \ge b \ge c$; these three positive numbers are the side lengths of a triangle if and only if $a < b + c$. Since the coefficients of $P(x)$ are nonnegative, we have $P(a) \ge P(b) \ge P(c) > 0$; thus, we need to check that $\sqrt[n]{P(a)} < \sqrt[n]{P(b)} + \sqrt[n]{P(c)}$.

Let $P(x) = p_n x^n + p_{n-1} x^{n-1} + \dots + p_0$. Denote $G(x) = P(x)/x^n$. Note that
$$G(a) = p_n + \frac{p_{n-1}}{a} + \dots + \frac{p_0}{a^n} \le p_n + \frac{p_{n-1}}{b} + \dots + \frac{p_0}{b^n} = G(b)$$
and similarly, $G(a) \le G(c)$. Therefore,
$$\sqrt[n]{P(a)} = a \sqrt[n]{G(a)} < (b + c) \sqrt[n]{G(a)} \le b \sqrt[n]{G(b)} + c \sqrt[n]{G(c)} = \sqrt[n]{P(b)} + \sqrt[n]{P(c)},$$
which is what was required to prove.