Let $u(x)$ be a smooth function from $\mathbb{R}$ to $\mathbb{R}$. Suppose that for some real numbers $a,b$ with $a < b$ the following equality is true:


\begin{equation}
\frac{1}{b-a} \int\_a^b u(x) \ \rm{d} x = \frac{1}{u(a)-u(b)} \int\_{u(b)}^{u(a)} x \ \rm{d} x.
\end{equation}


I would like to prove that, if $f$ is a non-negative convex function from $\mathbb{R}$ to $\mathbb{R}$, then the following inequality holds:


\begin{equation}
\frac{1}{b-a} \int\_a^b f(u(x)) \ \rm{d} x \geq \frac{1}{u(a)-u(b)} \int\_{u(b)}^{u(a)}f(x) \ \rm{d} x.
\end{equation}


$\bf{EDIT}$: It is clear now that, as written, the above claim is false. A related question is for what class of functions $f$ is the stated inequality true?



Integral inequality for convex function
The inequality is false in general. To see this, take $a=0,$ $b=1,$ $f(x)=x^2.$ Let $u(0)=0$
and $u(1)=1.$ Then our problem can be reformulated as follows: given that $\int\_{0}^1u(x)dx=\frac{1}{2}$ show that $\int\_{0}^1u^2(x)dx\ge\frac{1}{3}.$ Now take $u(x)=x+th(x)$ where $h(0)=h(1)=0$ and $\int\_{0}^1h(x)dx=0$ to end up with the inequality
$$t^2\int\_{0}^1h^2(x)dx+2t\int\_0^1xh(x)dx\ge 0$$ for all $t\in\mathbb{R}.$
Taking $t$ sufficiently small we get $\int\_0^1xh(x)dx\ge 0$ for all appropriately chosen $h(x).$ It is now easy to choose $h$ in way that last inequality is false (just take something antisymmetric with respect to $x=\frac{1}{2}$). 


