How do I show the following integral inequality?

Question

I have the next exercise.

Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Show that $\int_{0}^{1}f^2(t)dt\geq(\int_{0}^{1}f(t)dt)^2$.

I know it can be done using Cauchy-Bunyakovsky-Schwarz inequality, but can is there any other way to solve it?

you can use also the Hölder inequality, where the Cauchy-Schwarz inequality can be seen as a particular case. — Masacroso, Nov 25 '17 at 18:42

score 5 · Accepted Answer · answered Nov 25 '17 at 18:42

5

Hint. Note that for any real $x$, $$0\leq \int_{0}^{1}(x+f(t))^2dt=x^2+2x\int_{0}^{1}f(t)dt+\int_{0}^{1}f^2(t)dt$$ Now let $x=-\int_{0}^{1}f(t)dt$.

answered Nov 25 '17 at 18:42

Robert Z

why is that integral positive? – Ghost Nov 25 '17 at 18:57
It is the integral of a non negative function (a function squared). – Robert Z Nov 25 '17 at 18:58
Ok I think it works but just let me know how did you get this idea? – Ghost Nov 25 '17 at 19:22
@OvyOvy Start with a "trivial" (integral) inequality with a parameter and then play with the parameter is common strategy for this kind of stuff. See for example here for a similar approach: https://math.stackexchange.com/questions/492294/how-prove-this-integral-inequality?rq=1 – Robert Z Nov 25 '17 at 19:26

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