Proof Expected utility vs Utility of expected value

Question

I have a feeling it is a trivial question, but I cannot see how to prove it.

Let $U(\mathbb{E}(p)) > U(\mathbb{E}(q))$,

where $p$ and $q$ are probability distribution

I need to show that $\mathbb{E}[U_p] > \mathbb{E}[U_q]$, where the last expression can be written as $\int u(x)p(x)dx > \int u(x)q(x)dx$

How could I approach this problem?

Edit. Further assumptions

1. $F(x) < G(x)$ for every $x$, where $F()$ and $G()$ are the cdf of $p$ and $q$, respectively.
2. $U()$ increasing

Answer 1

$$ \hspace{-1.7cm} F(x) < G(x) $$

u(x) is increasing:

$ \begin{align*} \implies-u'(x)F(x)&>-u'(x)G(x) &\text{}\\ \implies-\int_{-\infty}^{+\infty}u'(x)F(x)&>-\int_{-\infty}^{+\infty}u'(x)G(x) \\ \implies[u(x)F(x)]_{-\infty}^{+\infty}-\int_{-\infty}^{+\infty}u'(x)F(x)&>[u(x)G(x)]_{-\infty}^{+\infty}-\int_{-\infty}^{+\infty}u'(x)G(x)\\ \end{align*}$

From integration by parts:

$\begin{align*} \hspace{3.1cm}\implies\int_{-\infty}^{+\infty}u(x)dF(x)&>\int_{-\infty}^{+\infty}u(x)dG(x)&\text{}\\ \implies\mathbb{E}(U_p)&>\mathbb{E}(U_q) \end{align*} $

I think there should be an additional assumption that p and q have common support and $u(x)$ is well defined in that support.