Strategy to prove $\int_0^1(f'(x)^2dx \geq3 (\int_0^1 f(x)dx)^2 $ if $f(1)=0$

Question

$f$ is differentiable, derivative continuous, $f(1)=0$. What is the strategy to prove: $\int^1_0(f'(x)^2dx \geq 3(\int^1_0 f(x)dx)^2$?

Please do not post a solution. Instead, help me to discover the correct approach to solving the problem. Below is what I have so far; please critique it, refine it, encourage it, discourage it, probe it, or add a new angle.

Let $k = -f(0)$. By the Mean Value Theorem, there must exist $c \in (0,1)$ such that $f'(c) = k$. Assume for the moment that $f'$ is constant across $(0,1)$. Then the left side equals $k^2$, and the right $3k^2/4$.

Of course, $f'$ need not be constant. It may jiggle higher or lower than $k$. However, $\int^1_0f'(x)dx$ must still equal $k$, by the Fundamental Theorem of Calculus. Intuitively, the "weighted average" of $f'$ is fixed by the endpoints of the interval, even if the value at a given point across the interval is not.

My intuition is that any such jiggling of $f'$ will raise the left side (the sum of the squares) faster that it raises the right side (the square of the sums). This is suggested (not proven) by: Given $b+c=0$, $(a^2 + b^2 + c^2) \geq (a+b+c)^2$. However, I do not know how to turn this bit of intuition into a proof. I don't even know if this intuition is actually correct!

I've tried other ways to use the Fundamental Theorem of Calculus to relate the two sides of the inequality, but to no avail. I've also tried using standard integration techniques to transform either side (e.g. using integration by change of variables, treating the left side as $f' \circ x \mapsto x^2$) without progress either.

Answer 1

You're on the right track. Remember that the Schwartz inequality for integrals reads

$$\left(\int_{0}^1 A(x)B(x)dx\right)^2\leq \int_0^1 A^2(x)dx\int_0^1 B^2(x)dx$$

All you have to do is choose the right functions to apply the inequality on. It's clear that one of them has to be equal to $f'$ to reproduce the RHS, but what is the other? HINT: find a simple function which, when squared, has an integral over the interval $(0,1)$ equal to $1/3$ and provides a simplification for the LHS.

Solution: set $A(x)=x, B(x)=f'(x)$ and note that $\int_{0}^1 x f'(x)dx=-\int_0^1 f(x)dx$.