Proof: Risk averse; Certainty Equivalent smaller than expected value I would like to show for a randomly distributed variable $x$ with CDF $F(\cdot)$ , given a Bernoulli utility function $u(x)$ the following property holds:

The certainty equivalent, $CE(\cdot)$, is smaller than the expected value, $\mathbb E(\cdot)$, if and only if the decision maker is risk-averse. 

$$CE(F,u) \leq \int_{-\infty}^{\infty} x dF(x) \quad  \forall F(\cdot)$$ 

By Jensens Inequality for risk-averse agents we have: 
$$\int_{-\infty}^{\infty} u(x) f(x)dx \leq u(\int_{-\infty}^{\infty} x f(x) dx)$$

I know that the CE is defined as: 
$u(CE) = \mathbb E(U(X)) = \int_{-\infty}^{\infty} f(x)u(x)\,dx$ 

Hence, 

$$u(CE)  \leq u(\int_{-\infty}^{\infty} x f(x) dx)$$ 

and

$$CE \leq \int_{-\infty}^{\infty} x f(x) dx$$

Is it possible to kind of invert the u() function is we assume it is strictly increasing?

I also thought about rearranging it to: 

$$u(CE) = \int_{-\infty}^{\infty} f(x) dx * \int_{-\infty}^{\infty} u(x) dx$$

but I fail to see how that helps.

I thought perhaps showing that the risk premium is positive for risk-averse agents is equivalent but I could not get started either. 

I would be glad for a hint to work out the solution myself.
You basically have the proof, just remember that for any function $f(x)$ strictly increasing, then $f(a)\leq f(b)$ if and only if $a\leq b$. It is also true that for any strictly increasing function its inverse exists, but there is probably no need for that.