Q:

Comparing $\int_{0}^{1}f(t)^2 dt$ to $\int_{0}^{1}f'(t)^{2} dt$ when $f(0)=0$

This is from an MCQ contest.

Let $f: [0,1]\to \mathbb{R}$ be a function of class $C^{1}$ with $f(0)=0$.
Which inequality is true?

$1]$ ${\displaystyle \int_{0}^{1}\bigl[f(t)\bigr]^{2} dt \leq \dfrac{1}{4}\int_{0}^{1}\bigl[f'(t)\bigr]^{2} dt }$
$2]$ ${\displaystyle \int_{0}^{1}\bigl[f(t)\bigr]^{2} dt \leq \dfrac{1}{3}\int_{0}^{1}\bigl[f'(t)\bigr]^{2} dt }$
$3]$ ${\displaystyle \int_{0}^{1}\bigl[f(t)\bigr]^{2} dt \leq \dfrac{1}{2}\int_{0}^{1}\bigl[f'(t)\bigr]^{2} dt }$
$4]$ ${\displaystyle \int_{0}^{1}\bigl[f(t)\bigr]^{2} dt \geq \int_{0}^{1}\bigl[f'(t)\bigr]^{2} dt }$

My Thoughts
note that ${\displaystyle  f(t)^2 =\left( \int_{0}^t f'(s)ds\right)^2 }$ 
$$\int_0^1 |f(t)|^2dt =  \int_0^1  \left( \int_{0}^t f'(s)ds \right)^2dt $$ 
With Cauchy Schwarz:
$$ \int_0^1 f'(s)^2 $$
Since this was taken from math-Contest MCQ I'm interested to know the fastest way to solve it 

A:

By the fundamental theorem of calculus, since $f(0)=0$,
$$
f(t)=\int_0^tf'(s)\,ds.
$$
Hence, by Cauchy-Scwarz inequality, and domain monotonicity of integrals of non-negative functions,
$$
\begin{aligned}
\int_0^1 (f(t))^2\,dt&=\int_0^1\biggl[\int_0^t 1\cdot f'(s)\,ds\biggr]^2\,dt\\
&\leq \int_0^1\biggl[\int_0^t1^2\,ds\int_0^t (f'(s))^2\,ds\biggr]\\
&\leq \int_0^1 t \,dt\int_0^1 (f'(s))^2\,ds\\
&=\frac{1}{2}\int_0^1 (f'(s))^2\,ds.
\end{aligned}
$$
This implies that 3] is true. As far as discussed in the comments, the other ones are not.
PS If I'm not mistaken, the $1/2$ in this inequality could be improved (using the minimization characterization of eigenvalues of $-D^2$) to be one divided by the minimum of the function
$$
\lambda\mapsto \frac{4\lambda^{3/2}}{2\sqrt{\lambda}-\sin(2\sqrt{\lambda})}-\lambda
$$
which is something like $2.467$, i.e. the $1/2$ can be replaced by $1/2.467\approx 0.4053$
Edit Explanation of the quotient. The quotient
$$
\frac{\int_0^1(f'(t))^2\,dt}{\int_0^1 (f(t))^2\,dt}
$$
is the Rayleigh quotient for the operator $-D^2$ on the interval $[0,1]$. So, if we assume that we have a self-adjoint realization of $-D^2$ then the minimum of this quotient will correspond to the lowest eigenvalue of that operator, when minimized over the right domain. Now, solving
$$
-D^2u=\lambda u,\quad u(0)=0
$$
gives
$$
u=C\sin(\sqrt{\lambda}t).
$$
Calculating the Rayleigh quotient of this function, one gets the quotient I mentioned in the PS. Now we need more conditions for the operator to be self-adjoint (typically one at $t=1$). But minimizing the quotient we find a certain $\lambda$, and that $\lambda$ gives a condition at $t=1$. With that condition we have (choosing domains correctly) a self-adjoint realization. Probably a bit more hand-waving is needed if one wants to be completely rigorous, but this is at least an explanation of how one could get that quotient.

