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?

Question

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.

Answer 1

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: $$ \begin{align} \sqrt[n]{P(a+b)} &= a \sqrt[n]{Q(a+b)} + b \sqrt[n]{Q(a+b)} \\ &\le a \sqrt[n]{Q(a)} + b \sqrt[n]{Q(b)} \\ &= \sqrt[n]{P(a)} + \sqrt[n]{P(b)} \end{align} $$