Prove that $\int_{0}^{1}f(x)^2dx\geq 4$ Let $f:[0,1]\rightarrow \mathbb{R} $ be an integrable function with $\int_{0}^{1}f(x)dx=\int_{0}^{1}xf(x)dx=1$.
Prove that $\int_{0}^{1}f(x)^2dx\geq 4$.
I got that $\int_{0}^{1}F(x)dx=F(0)$, but I don't think it's useful at all.
 A: Observe that $\int_{0}^{1}f(x)g(x)\mathrm{d}x$ defines an inner product on the space of square-integrable functions on $[0,1].$ If $\int_{0}^{1}f(x)^{2}\mathrm{d}x=+\infty,$ then it is certainly at least 4, so we may restrict to this case. Now observe that if $f(x)=a+bx,$ then $\int_{0}^{1}f(x)^{2}\mathrm{d}x=a^{2}+ab+b^{2}/3,$ and $$\int_{0}^{1}f(x)\mathrm{d}x=a+b/2,\quad\int_{0}^{1}xf(x)\mathrm{d}x=a/2+b/3.$$ Now if $f$ is a general square-integrable function, it must be of the form $a+bx+g(x),$ where $g(x)$ is orthogonal to $1$ and $x$ with respect to this inner product, i.e., $\int_{0}^{1}g(x)\mathrm{d}x=\int_{0}^{1}xg(x)\mathrm{d}x=0.$ Then we have that \begin{align*}\int_{0}^{1}f(x)^{2}\mathrm{d}x&=\int_{0}^{1}(a+bx)^{2}+2(a+bx)g(x)+g(x)^{2}\mathrm{d}x\\&=\int_{0}^{1}(a+bx)^{2}+g(x)^{2}\mathrm{d}x\\&\geq \int_{0}^{1}(a+bx)^{2}\mathrm{d}x=a^{2}+ab+b^{2}/3.\end{align*}
Solving the system of equations $a+b/2=1,$ $a/2+b/3=1,$ we get that $a=-2$ and $b=6,$ so we have that $\int_{0}^{1}f(x)^{2}\mathrm{d}x\geq 4.$ 
A: I would begin with
$$\int_0^1 dx \left [f(x) - (a x+b) \right ]^2 \ge 0 $$
for any $a,b \in \mathbb{R}$.  Expand the integrals and use the info provided to get that
$$\int_0^1 dx \, f(x)^2 \ge 2 a+2 b - \frac13 a^2-a b-b^2 $$
I would then maximize the expression on the right.  Let the expression on the RHS be $g(a,b)$.
We may maximize by taking derivatives of $g(a,b)$ wrt $a$ and $b$ and setting them equal to zero.   We then get a system of equations
$$2 a + 3 b = 6$$
$$a+2 b=2$$
which implies that $a=6$ and $b=-2$. Note that $g_{aa} g_{bb} - g_{ab}^2 = \frac13 \gt 0$ and $g_{aa} = -\frac23 \lt 0$, so the critical point we found is indeed a maximum.  It is then easy to verify that $g(6,-2)=4$.
A: $Michael 's idea:
$(a+b)^2=\left(\int_0^1(ax+b)f(x)dx\right)^2=T$
We know Cauchy-Schwarz inequality:
$\left(\int_Df\cdot gdx\right)^2\le \int_Df^2dx\int_Dg^2dx$
We use this inequality:
$(a+b)^2=T\le \int_0^1(f(x))^2dx\int_0^1(ax+b)^2dx$
$\int_0^1(f(x))^2dx\ge \frac{(a+b)^2}{\frac{a^2}{3}+ab+b^2}$
$a=-3b$
