# Prove the following integral inequality

Suppose $f(x)$ and $g(x)$ are continuous function from $[0,1]\rightarrow [0,1]$, and $f$ is monotone increasing, then how to prove the following inequality: $$\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx$$

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I'll be the first to say it - users are encouraged to say what they've done on a question so far/what might be a point of confusion so that people can deliver more helpful (and not redundant) advice. –  lamb_da_calculus Jan 13 '13 at 4:20
yes, because it's hard and I have no idea to go any further. I just get a hint: how to estimate $f(x)$, then change $x$ to $g(x)$ –  Larry Eppes Jan 13 '13 at 4:24
Did you consider using the fact that $f$ as a continuous monotone function is differentiable almost everywhere? –  Eckhard Jan 13 '13 at 15:44
$\int_{0}^{1}[f(g(x))-g(x)]dx=f(g(\xi))-g(\xi)=f(u)-u\leq f(u)-uf(u)=(1-u)f(u) =\int_{u}^{1}f(u)dx\leq\int_{u}^{1}f(x)dx\leq\int_{0}^{1}f(x)dx.$
Here we use the Mean value theorems for integration in the first $"="$, and let $u=g(\xi)$.