Inequality of integrals $\int_0^1(f(x))^2 dx \geq 4$ if $\int_0^1xf(x) dx=\int_0^1f(x) dx = 1$ If 
$$\int_0^1xf(x) dx=\int_0^1f(x) dx = 1$$
prove that
$$\int_0^1(f(x))^2 dx \geq 4$$
EDIT
My attempt is as follows - 
I can use only the $\int xf(x)$dx part to get a bound $\int f^2(x) dx \geq 3$ from cauchy schwarz. I can't think of ways how to incorporate the other given condition.
 A: Write your function in terms of the shifted Legendre polynomials $\tilde{L}_n(x)=L_n(2x-1)$, that are an orthogonal base of $L^2((0,1))$ with respect to the usual inner product. Assuming:
$$ f(x) = \sum_{n=0}^{+\infty} a_n\,\tilde{L}_n(x), $$
the constraints give:
$$ a_0 = 1,\qquad \frac{a_0}{2}+\frac{a_1}{6} = 1, $$
hence $a_0=1$ and $a_1=3$. This implies:

$$\int_{0}^{1}f(x)^2 dx = a_0^2+\sum_{n=1}^{+\infty}\frac{a_n^2}{2n+1}\geq 1+\frac{9}{3}=4.$$

Moreover, you have that equality holds only for $f(x)=6x-2$.
A: We have for each $a,b$:
$$ \int_0^1 \Big(f(x)-ax-b\Big)^2dx \geq 0$$
So we have
\begin{eqnarray*}
\int_0^1 f(x)^2 dx &\geq &2\int _0^1f(x)(ax+b)dx -\int _0^1
(ax+b)^2dx \\
&=&2(a+b) -{(a+b)^3-b^3\over 3a}\\
&=&2(a+b) -{a^2+3ab+3b^2\over 3} =:E
\end{eqnarray*}
This inequality is valid for all $a,b$, so even when $E$ achieves maximum:
\begin{eqnarray*}
E &=&\underbrace{-b^2 +b(2-a) - {(2-a)^2\over 4}} + \underbrace{{(2-a)^2\over
4} -{a^2\over
3}}\\
&=&-\Big(b - {2-a\over 2}\Big)^2+{-a^2-12a+12\over 12}\\
    &\leq & {-a^2-12a+12\over 12} \\
   &=&  {-(a-6)^2+48\over 12}\\
    &\leq & 4
\end{eqnarray*}
Equality is achieved at $a=6$ and $b={2-a\over 2}= -2$, that is when $f(x)=6x-2$.
A: Of course once the inequality condition is obtained (or guessed by trying out a general linear polynomial), Cauchy-Schwarz is a breeze:
$$\int_0^1 f^2 dx \ge \frac{\left(\int_0^1 (3xf-f)dx\right)^2}{\int_0^1(3x-1)^2 dx } = \frac{4}{1}=4$$
Equality is when $f(x)$ is proportional to $3x-1$.
In general we can have
$$\int_0^1 f^2 dx \ge \frac{\left(\int_0^1 (axf+bf)dx\right)^2}{\int_0^1(ax+b)^2 dx } = \frac{(a+b)^2}{a^2/3+ab+b^2}$$
where the maximum is when $(a, b)$ is proportional to $(3, -1)$, so $4$ is indeed the best possible.
