How to prove Godunova's inequality? 
Let $\phi$ be a positive and convex function on $(0,\infty)$. Then
  $$\int_0^\infty \phi\left(\frac{1}{x}\int_0^x g(t)\,dt\right)\frac{dx}{x} \leq \int_0^\infty \phi(g(x))\frac{dx}{x}$$

The application of this inequality is this :
$(1)$ Hardy's inequality. 
With $\phi(u)=u^p$, we obtain that
$$\int_0^\infty \left(\frac{1}{x} \int_0^x g(t) \, dt\right)^p\frac{dx}{x} \leq \int_0^\infty g^p(x)\frac{dx}{x}$$
(2) Polya-Knopp's inequality
By using it with $\phi(u)=u^p$, replacing $g(x)$ by $\log g(x)$ and making the substitution $h(x)=\frac{g(x)}{x}$ we obtain that
$$\int_0^\infty \exp\left(\frac{1}{x} \int_0^x \log h(t)\right) \, dt \leq e\int_0^\infty h(x) \, dx$$
How to prove Godunova's inequality? Is there any reference?

This is just an explicit execution (as a community wiki answer) of the hint given in the comments:
By Jensen's inequality, we get (because $(0,x)$ with the measure $\frac{dt}{x}$ is a probability space)
\begin{eqnarray*}
\int_{0}^{\infty}\phi\left(\int_{0}^{x}g\left(t\right)\frac{dt}{x}\right)\frac{dx}{x} & \leq & \int_{0}^{\infty}\int_{0}^{x}\phi\left(g\left(t\right)\right)\frac{dt}{x}\,\frac{dx}{x}\\
 & \overset{\text{Fubini}}{=} & \int_{0}^{\infty}\phi\left(g\left(t\right)\right)\,\int_{t}^{\infty}x^{-2}\, dx\, dt\\
 & = & \int_{0}^{\infty}\phi\left(g\left(t\right)\right)\,\frac{dt}{t}
\end{eqnarray*}
The application of Fubini's theorem is legitimate as the integrand is non-negative.
