If $$P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0\, ,$$then by Minkowski inequality \begin{align*} \sqrt[n]{P(a)}+ \sqrt[n]{P(b)} \geq \sqrt[n]{\sum_{k=0}^{n} a_k\left(a^{\frac{k}{n}} + b^{\frac{k}{n}}\right)^{n} } \geq \sqrt[n]{\sum_{k=0}^{n} a_k\left(a + b\right)^{k} } = \sqrt[n]{P(a+b)} > \sqrt[n]{P(c)} \end{align*}

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0,$ and $Q(x)=\frac{P(x)}{x^n}.$ We know $a<b+c.$ $(1)$ Wlog: $a\ge b\ge c,$ then we have $Q(a)=a_n+\frac{a_{n-1}}{a}+...+\frac{a_0}{a^n}\le a_n+\frac{a_{n-1}}{b}+...+\frac{a_0}{b^n}=Q(b).$ Similarly $Q(b)\le Q(c),$ then $Q(a)\le Q(b)\le Q(c).$ We must prove that $$\sqrt[n]{P(a)}=a\sqrt[n]{Q(a)}<^{(1)}b\sqrt[n]{Q(a)}+c\sqrt[n]{Q(a)}\le b\sqrt[n]{Q(b)}+c\sqrt[n]{Q(c)}=\sqrt[n]{P(b)}+\sqrt[n]{P(c)}.$$As desired.

decreasing function

$P(x)$ is polynomial with degree $n\geq 2$ and nonnegative coefficients. $a,b,c$ - sides for some acute triangle. Prove, that $\sqrt[n]{P(a)},\sqrt[n]{P(b)},\sqrt[n]{P(c)}$ are sides for some acute triangle too.