Prove intergral inequality

Question

If $f$ is a Riemann-integrable function on $[a,b]$ for which $\int\limits_a^b f(x) dx = 0$, and $m \leq f(x) \leq M$ for all $a \leq x \leq b$, then prove that $$\int\limits_a^b f(x)^2 dx \leq - m M (b-a).$$

My only idea how to use $\int\limits_a^b f(x) dx = 0$ is $$\int\limits_a^b f(x)^2 dx = \int\limits_a^b f(x)(f(x)+A) dx,$$ with some constant $A$. And try to use Cauchy-Schwarz inequality $$\int\limits_a^b f(x)(f(x)+A) dx \leq \sqrt{\int\limits_a^b f(x)^2 dx} \sqrt{\int\limits_a^b (f(x)+A)^2 dx},$$ that could lead to $$\int\limits_a^b f(x)^2 dx \leq \max\{-m, M\} (M-m) (b-a)$$ with appropriate $A$. But this inequality is much weaker than the one I need.

I would appreciate any tips regarding this problem. Thanks!

Solved

If anyone is interested, here is the solution: $$m \leq f(x) \leq M => (f(x) - m)(M - f(x)) \geq 0,$$ $$\int\limits_a^b (f(x) - m)(M - f(x))dx \geq 0,$$ $$\int\limits_a^b \left(- f^2(x) + (M + m)f(x) - m M \right)dx \geq 0,$$ $$\int\limits_a^b f^2(x)dx \leq - m M (b-a)$$

if m and M have the same signs, then it seems that the inequality does not hold. — Sara, Aug 25 '19 at 15:57
If $m$ and $M$ have the same sign, what must they be for $\int_a^bf(x),\mathrm{d}x=0$? — robjohn, Aug 25 '19 at 16:52

score 1 · Answer 1 · answered Aug 15 '15 at 08:18

It seems that your solution is correct. Nevertheless, I think that some words need to be added between each displayed equation. For example, between the second and third equation, you should say something like "Expanding the integrand, we have..." and between the third and fourth, you should say that you used the fact that $\int_a^bf(x)\mathrm dx$.

Prove intergral inequality

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