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

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!

This is not an answer, @Rienmann provided one. But it is an interesting corollary of this result:

$$\int_0^1 f(f(x)) dx \leq 2\int_0^1 f(x)dx$$

Also,

$$\int_0^1 f(x) dx + \int_0^1 f^{-1}(x) dx \geq 1$$

and

$$\int_0^1 f(x^2) dx \leq \frac{1}{3}+\int_0^1 f(x) dx.$$ An interesting question is when (for which functions) the last inequality becomes an equality.