Prove the following integral inequality: $\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx$

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

By the Mean Value Theorem for Integrals there is a point $\xi$ in $[0,1]$ such that $$\int_{0}^{1}[f(g(x))-g(x)]dx=f(g(\xi))-g(\xi)$$ Let $u = g(\xi)$. Then $$f(g(\xi))-g(\xi) = f(u)-u \leq f(u) - uf(u) = (1-u)f(u)$$ (the inequality is due to the fact that $0 \leq f(u) \leq 1$.) Since $f$ is monotone increasing, $f(x) \geq f(u)$ for all $x$ in $[u,1]$. So $$(1-u)f(u) = \int_u^1 f(u)\,dx \leq \int_u^1 f(x)\,dx \leq \int_0^1 f(x)\,dx$$ Rearranging, $$\int_0^1 f(g(x)) \,dx \leq \int_0^1 f(x)\,dx + \int_0^1 g(x)\,dx$$ Over!