Property of integration Let $T>0$ fixed. Let $f,g : [0,T] \subset \mathbb{R} \longrightarrow \mathbb{R}$ differentiable functions such that, for some $c>1$,
$$f(x) \geq c, \; \forall \; x \in [0,T] .$$
I know that $f\cdot g$ is also integrable.
Question. Is it true that
$$\int_{0}^T c \cdot g(x) \; dx \leq \int_{0}^T f(x) \cdot g(x) \; dx?$$
I thought of applying integration by parts to try to prove it, but I could not conclude anything.

This is false. $f(x)=c+1$ and $g(x)=-1$ gives a  counter-example.
However, if we also have $g(x) \geq 0$ then $f(x)g(x)-cg(x) \geq 0$ so $\int_0^{T} [f(x)g(x)-cg(x)] \geq 0$ which gives $\int_0^{T} f(x)g(x) \geq c \int_0^{T} f(x)dx$.
