Let $f,g:[a,b] \longrightarrow \mathbb{R}$ such that $f$ is continuous and $p$ integrable with $p(x) > 0$ for all $x \in [a,b]$. Prove that if $$\int\limits_{a}^{b}f(x)p(x)\mathrm{d}x = f(a)\int\limits_{a}^{b}p(x)\mathrm{d}x$$ then there exists $c \in (a,b)$ such that $f(c) = f(a)$.
I tried assume that $f(x) > f(a)$ for all $x \in (a,b)$, after, assume that $f(x) < f(a)$ for all $x \in (a,b)$. But applying the integral in $f(x)p(x) > f(a)p(x)$ and $f(x)p(x) < f(a)p(x)$, I'll get non-strict inequalities.
Any idea?
If $\displaystyle\int_a^bg(x)p(x)\,dx=0$ with $g$ continuous, you get from $p>0$ that there exists $c\in (a,b)$ with $g(c)=0$. Otherwise, the integral cannot be zero.
Now take $g(x)=f(x)-f(a)$.