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Let $f$ be a continuously differentiable real valued function on $[0,1]$.

It is given that $displaystyle int_{frac{1}{3}}^{frac{2}{3}}f(x) dx=0$

Find the minimum value of $dfrac{int_{0}^{1} (f'(x))^2 dx}{left( int_{0}^{1} f(x) dx right)^2}$

I tried to use Cauchy-Schwarz to show that
$$frac{int_{0}^{1} (f'(x))^2 dx}{left( int_0^1 f(x) dx right)^2} ge frac{left( int_0^1 bigl| f(x)f'(x) bigr| dx right)^2}{ left( int_0^1 f(x) dx right)^2} ge frac{ f(1)^2 - f(0)^2}{2 left( int_0^1 f^2 (x) dx right)^2}$$
But I can't proceed from here.

Also, I don't know how to use the condition $int_{1/3}^{2/3}f(x) dx=0$

The minimum ratio is $27$, achieved by a piecewise quadratic polynomial in $x$.

This is sort of expected. When one throw the functional

$$int_0^1 f'(x)^2 dx$$
to Euler-Lagrange equation and subject it to the constraints
$$int_0^1 f(x) dx = text{constant}quadtext{ and }quad
int_{1/3}^{2/3} f(x) dx = 0$$
One find $f''(x)$ has to be a piecewise constant.
It takes one value over $[0,frac13) cup (frac23,1]$ and another value over $(frac13,frac23)$. What I has done is use a CAS to pin down the correct piecewise quadratic polynomial and then verify it give us the minimum.

Let $X = mathcal{C}^1[0,1]$ and $P,Q,C : X to mathbb{R}$ be the functionals over $X$ defined by

$$P(f) = int_0^1 f'(x)^2 dx,quad
Q(f) = int_0^1 f(x) dx,quadtext{ and }quad C(f) = int_{1/3}^{2/3} f(x) dx$$

The question can be rephrased as

What is the minimum of the ratio $frac{P(f)}{Q(f)^2}$ for $f in X$ subject to the constraint $C(f) = 0$.

Since the ratio and constraint are both invariant under scaling of $f$ by constant, we
can restrict our attention to those $f$ which satisfies $C(f) = 0$ and $Q(f)$ equal to a specific constant.

For any $K in mathbb{R}$, let $Y_K = big{; f in X : C(f) = 0, Q(f) = K; big}$.

Consider following function over $[0,1]$

$$g(x) = begin{cases}
4 - 27x^2, & x in [0,frac13]\
54(x^2-x) + 13, & x in [frac13,frac23]\
4 - 27(1-x)^2, & x in [frac23,1]
end{cases}$$
We have
$$
g'(x) = begin{cases}
-54x, & x in [0,frac13]\
54(2x-1),& x in [frac13,frac23]\
54(1-x),& x in [frac23,1]
end{cases},quad
g''(x) = begin{cases}
-54, & x in [0,frac13) cup (frac23,1]\
108, & x in (frac13,frac23)
end{cases}
$$
It is not hard to see $g in X$. With a little bit of effort, one can verify
$P(g) = 108$, $Q(g) = 2$ and $C(g) = 0$. This means $g in Y_2$.

For other $f in Y_2$, it is easy to see $eta = f - g in Y_0$. We can decompose $P(f)$ as follows

$$P(f) = int_0^1 (g'(x)+eta'(x))^2 dx = int_0^1 (g'(x)^2 + eta'(x)^2 + 2g'(x)eta'(x)) dx$$
Let us look at the cross term. Integrate by part and using the fact $g'(0) = g'(1) = 0$, we find

$$begin{align}int_0^1 g'(x)eta'(x) dx
&= [ g'(x) eta(x) ]_0^1 - int_0^1 g''(x)eta(x) dx\
&= 54 int_0^1 eta(x) dx - 162int_{1/3}^{2/3}eta(x)dx\
&= 54 Q(eta) - 162 C(eta)
end{align}
$$
Since $eta in Y_0$, $Q(eta) = C(eta) = 0$ and the cross term goes away. As a result,

$$P(f) = P(g) + P(eta) ge P(g)$$
because $P(eta)$ is non-negative. Together with $Q(f) = Q(g) = 2$, we get

$$frac{P(f)}{Q(f)^2} ge frac{P(g)}{Q(g)^2} = frac{108}{2^2} = 27$$

As a result,

$$minleft{ frac{P(f)}{Q(f)^2} : f in X, C(f) = 0 right} =
minleft{ frac{P(f)}{Q(f)^2} : f in Y_2 right} = 27$$