Hello!

I have the following problem:

Let $\displaystyle g:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable function with increasing derivative. And let $\displaystyle (X,M,\mu)$ be a finite measure space such that $\displaystyle \mu(X)=1$.

Prove: $\displaystyle f$ bounded and measurable then the following holds

$\displaystyle g(\int_Xfd\mu)\le \int_X g(f)d\mu<\infty$

The hint is to use the fact that $\displaystyle g$ lies above any of its tangent lines.

------------------------------------------------------------------

Since $\displaystyle g'$ is increasing it is easy to prove that $\displaystyle g\ge g(y_0)+g'(y_0)(y-y_0)$. I tried to use it in the following way - but without result:

$\displaystyle \int_Xg(f(x))d\mu(x)\ge\int_X(g(f(x_0))+g'(f(x_0)) (f(x)-f(x_0)))d\mu(x)=g(f(x_0))+g'(f(x_0))\int_X(f(x)-f(x_0))d\mu(x)=g(f(x_0))-g'(f(x_0))f(x_0)+g'(f(x_0))\int_Xfd\mu$

What am I missing?