$$\sum_{j=1}^{k}P(x_j)\geq k P\left(\frac{x_1+x_2+\dots+x_k}{k}\right)\geq kP(1)$$because $P$ is convex and increasing on $\mathbb{R}_{\geq 0}$.

Let $P(x)=\sum_{0\le j\le d}\alpha_j x^j$ where $d={\rm deg}(P)$. Now, \[ \sum_{1\le i\le k}P(x_i) = \sum_{0\le j\le d}\alpha_j\sum_{1\le i\le k} x_i^j\ge \sum_{0\le j\le d}\alpha_j k\left(\prod_{1\le i\le k}x_i\right)^{\frac{j}{k}}=k\sum_{0\le j\le d}\alpha_j = kP(1), \]where the inequality step uses the AM-GM inequality.

Jensens gives that $\frac{\sum P(x_i)}{k}\ge P(\frac{\sum x_i}{k})$ as $P$ is convex over the positive reals given its non-negative coefficients. By AM-GM we also have that $\frac{\sum x_i}{k}\ge 1$ and we are done as $P$ is also increasing over the positive reals.

Solution only using AM-GM. By AM-GM $x_1^n+x_2^n...+x_k^n \geq k \sqrt[k]{(x_1x_2...x_k)^n}=k$ If we let $P(x)=a_mx^m+a_{m-1}x^{m-1}+...+a_0$ then, $P(x_1)+P(x_2)+...+P(x_k)$ $=a_m(x_1^m+x_2^m+...+x_k^m)+...+a_1(x_1+x_2+...+x_k)+ka_0 $ $\geq a_m(k)+a_{m-1}(k)+...+a_0(k)$ $=kP(1)$.

This since we know the coefficient are non negative and therefore the second derivative of the polynomial is always greater than 0 there then applying jensen we get the results

If $P$ and $Q$ satisfy the desired claim, so do $P + Q$ and $aP$ for any $a\geq 0$, thus it is enough to check it for $P(x) = x^m$ and in this case it follows directly by AM-GM.