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I'm in the midst of studying for my real analysis final tomorrow and I came across this question in a practice final and am stumped on how to even start it.

Let $f: [0,1] \rightarrow \mathbb{R}$ be a differentiable function such that $f'$ is continuous and $f(1)=0$. Prove that the following inequality holds: $\int_{0}^{1}(f'(t))^2dt \geq 3(\int_{0}^{1}f(t)dt)^2$.

Any guidance on how to start would be of help..

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    $\begingroup$ I'm not sure but I would have a look over the Cauchy Schwarz inequality $\endgroup$ – Stu Aug 9 '17 at 23:34
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First apply Integration by Parts:

$$\int_0^1f(x)\;dx=-\int_0^1xf'(x)\;dx$$

Here we have used the fact that $xf(x)$ vanishes at both $0$ and $1$.

Now invoke Cauchy-Schwarz. We have $$\left(\int_0^1f(x)\;dx\right)^2=\left(\int_0^1xf'(x)\;dx\right)^2≤\int_0^1\left(f'(x)\right)^2\;dx\;\times\;\int_0^1x^2\;dx$$

And your inequality follows at once.

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    $\begingroup$ And equality holds if and only if $f'$ is a scalar multiple of $x$, which combined with $f(1) = 0$ implies $f(x) = \lambda (1 - x^2)$. $\endgroup$ – Daniel Schepler Aug 10 '17 at 0:09
  • $\begingroup$ @lulu: Very nice! (+1) $\endgroup$ – Markus Scheuer Aug 10 '17 at 6:48

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