An inequality involving integrals Let be $f:[0,1] \longrightarrow R $, $f$ is an integrable function such that:
$$\int_{0}^{1} f(x) \space dx = \int_{0}^{1} xf(x) \space dx=1$$
I need to prove that:
$$\int_{0}^{1} f^2(x) \space dx\geq4$$
 A: Note that if $h(x)=-2+6x$ then $$\int_0^1 h(x)\, dx = \int_0^1 xh(x)\, dx =1.$$ Moreover 
$$
\int_0^1 (h-f)^2 dx\ge 0
$$
The rest is simple. Also Cauchy--Schwarz works with a slightly modified proof.  
A: A geometric reading of this question is to consider the space $E=\mathcal C([0,1],\mathbb R)$ with inner product  :  $$\langle f,g \rangle = \int_0^1  f(t) g(t) dt$$ and   say : Show that forall $f \in F^{\bot}$ , we  have : $\| f \|  \geq 2 $ where $F= \text{Span} \{x \mapsto 1, x \mapsto x \} $.
By Gram-Schmidt orthogonalization procedure, we obtain $(x \mapsto 1, x \mapsto \sqrt 3(2x-1))$ as orthonormal basis of $F$  who gives one expression of $p_F(f)$ projection of $f$. By Pythagore theorem we  have : $\|p_F(f)\|   \leq \|f\|$ and since the basis is orthonormal wa have : $$\|P_F(f)\|^2=(\langle f,1 \rangle )^2 + (\langle f,\sqrt 3(2x-1) \rangle )^2 = 1^2 + \sqrt 3^2 = 4$$
All this can explain the  provenace  of  the function $x \mapsto 6x-2 = 1 + 3(2x-1)  $ used by Unoqualunque.
Edit  :  $x \mapsto 6x-2 = 1 + 3(2x-1)  $ instead  of  $x \mapsto 6x-2 = 3(2x-1)  $
