Given such $n$th degree polynomial $P(x)$ and positive numbers $a, b\in\mathbb{R}$, does $\sqrt[n]{P(a+b)} \leq \sqrt[n]{P(a)} + \sqrt[n]{P(b)}$ hold? Given any $n$th degree polynomial $P(x)$ with positive coefficients and positive numbers $a, b\in\mathbb{R}$, does $\sqrt[n]{P(a+b)} \leq \sqrt[n]{P(a)} + \sqrt[n]{P(b)}$ hold?
I've worked it out for $n=2$:
$$P(x) = Ax^2 + Bx + C$$
$$\sqrt{Aa^2 + Ba + C} + \sqrt{Ab^2 + Bb + C} = \sqrt{(\sqrt{Aa^2 + Ba + C} + \sqrt{Ab^2 + Bb + C})^2} $$
$$= \sqrt{(Aa^2 + Ba + C) + 2\sqrt{(Aa^2 + Ba + C)(Ab^2 + Bb + C)} + (Ab^2 + Bb + C)}$$
$$ \geq \sqrt{Aa^2 + Ba + C + 2\sqrt{(Aa^2)(Ab^2)} + Ab^2 + Bb + C}$$
$$ \geq \sqrt{Aa^2 + Ba + C + 2Aab + Ab^2 + Bb}$$
$$ = \sqrt{A(a^2 + b^2 + 2ab) + B(a+b) + C} = \sqrt{A(a+b)^2 + B(a+b) + C}$$
But I couldn't figure out the generalization from there.
I also pondered whether Jensen's inequality would be of help, but I don't think $\sqrt[n]{P(x))}$ is concave.
I have used various higher degree polynomials to plot $\sqrt[n]{P(x))}$ and the inequality always seems to hold.

From https://artofproblemsolving.com/community/c6h1440215p8188094 on AoPS:
If $P(x) = \sum_{k=0}^n c_k x^k$ is a polynomial with nonnegative coefficients and $a, b \ge 0$ then
$$
\begin{align}
 \sqrt[n]{P(a+b)} &= \left( \sum_{k=0}^n c_k (a+b)^k\right)^{1/n}\\
 &\overset{(1)}{\le} \left(\sum_{k=0}^n c_k( a^{k/n} + b^{k/n})^n\right)^{1/n} \\
&\overset{(2)}{\le} \left(\sum_{k=0}^n  c_k a^k\right)^{1/n} + \left(\sum_{k=0}^n  c_k b^k\right)^{1/n}\\
&= \sqrt[n]{P(a)} + \sqrt[n]{P(b)}
\end{align}
$$
where

*

*$(1)$ follows from How to show that $(a+b)^p\leq a^p + b^p$ for $a,b\geqslant 0$ and $0<p<1$? with $p=k/n$, and

*$(2)$ is Minkowski's inequality.

Or, using the fact that $Q(x) = P(x)/x^n$ is decreasing:
$$
\begin{align}
\sqrt[n]{P(a+b)} &= a \sqrt[n]{Q(a+b)} + b \sqrt[n]{Q(a+b)} \\
 &\le a \sqrt[n]{Q(a)} + b \sqrt[n]{Q(b)} \\
 &= \sqrt[n]{P(a)} + \sqrt[n]{P(b)}
\end{align}
$$
