TITLE: Specific Integral Question: $\int_0^1(f'(t))^2dt \geq 3(\int_0^1f(t)dt)^2$
QUESTION [8 upvotes]: 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..

REPLY [13 votes]: 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.