1. ## An inequality

Hello!

I have the following problem:

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

Prove: $f$ bounded and measurable then the following holds
$g(\int_Xfd\mu)\le \int_X g(f)d\mu<\infty$

The hint is to use the fact that $g$ lies above any of its tangent lines.
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Since $g'$ is increasing it is easy to prove that $g\ge g(y_0)+g'(y_0)(y-y_0)$. I tried to use it in the following way - but without result:

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

What am I missing?

2. ## Re: An inequality

Apply the inequality $g(x)\geq g(y_0+g'(y_0)(x-x_0)$ to $x:=f(t)$ and $y_0:=\int_Xfd\mu$, then integrate.

3. ## Re: An inequality

And how do I show $\int_X g(f)d\mu<\infty$?

First I thought it had something to do with $f$ being bounded but $g$ is not the increasing one it's $g'$.

4. ## Re: An inequality

Since $f$ is bounded and $g$ continuous, $g\circ f$ is bounded. Now use the fact that we integrate over a finite-measure space.