Elementary proof of $f>0$ implies $\int f>0$?

The statement is true because one can show there exists $\xi \in (a,b)$ such that $$\int_a^b f(x) \, dx \geq f(\xi) (b-a)$$

In fact assume that $$f(x) > \frac 1{b-a} \int_a^b f(y)\,dy$$ for all $x \in (a,b)$

Changing the values of $f$ at $a$ and $b$ if necessary (which doesn't alter the value of the integral), we have a new function $\hat{f}$ such that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy$$ for all $x \in [a,b]$

Remember now the theorem discussed in this question.

Since $\hat{f}$ is continuous at some $c \in (a,b)$, we can find $\varepsilon > 0\,$ so that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy + \varepsilon$$ for all $x \in (c-\varepsilon,c+ \varepsilon) \subset (a,b)$.

Then, if we consider the partition $P=\{a,c-\varepsilon,c+\varepsilon,b\}$, we obtain $$\int_a^b \hat{f}(x)\,dx \geq L(\hat{f},P) \geq \int_a^b \hat{f}(x)\,dx + 2\varepsilon^2 > \int_a^b \hat{f}(x)\,dx$$ which is absurd.

The paper Rodrigo Lopez Pouso, Mean Value Integral Inequalities , Real Anal. Exchange Volume 37, Number 2, (2011), 439-450 is worth reading not only as the source of this proof.