Show that $\int_0^1 f^2(x)dx\geq 3$, if $f:[0,1]\to \mathbb{R}$ is an integrable function s.t. $\int_0^1 f(x)dx=\int_0^1 xf(x)dx=1$ So far I've done this, but I don't know if it will help.
Let $\int_0^xf(t)dt=F(x)$.
Now, $1=\int xf(x)=F(x)x|_0^1-\int_0^1F(x)dx=1-\int_0^1 F(x)dx$. Then, $\int_0^1 F(x)dx=0$.
But I don't know how to keep going.
 A: Hint: Use the Cauchy Schwarz inequality. That is,
$$
\left(\int f(x)g(x)\,dx\right)^2 \leq
\int f^2(x)\,dx \cdot \int g^2(x)\, dx
$$
A: The bound can be made a little sharper:
Since $f$ is integrable, we have $f \in L^2[0,1]$, so we can deal with the
Hilbert space $L^2[0,1]$.
Let $\phi_0(x) = 1$, then we have $\langle \phi_0 , f \rangle = 1$ and we
have $\|\phi_0\| = 1$.
Let $g_1(x) = x$, we have $\langle g_1 , f \rangle = 1$, but it is convenient to make $g_1$ orthogonal to $\phi_0$ and normalise. Using Gram Schmidt we
get $\phi_1(x) = \sqrt{12} (g_1(x) - {1 \over 2}) = \sqrt{12} (x - {1 \over 2}) $. Then $\langle \phi_1 , f \rangle = \sqrt{3}$ and we
have $\|\phi_1\| = 1$.
Then we have $f = \sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k + f-\sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k$ and since
$\sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k$ and $f-\sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k$ are orthogonal, we have
\begin{eqnarray}
\|f\|^2 &=& \|\sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k\|^2+\|f-\sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k\|^2 \\
&\ge& \|\sum_{k=0}^1 \langle \phi_k , f \rangle \phi_k\|^2 \\
&=& \sum_{k=0}^1 |\langle \phi_k , f \rangle|^2 \\
&=& 4
\end{eqnarray}
From which we get $\int_0^1 f^2(x)dx = \|f\|^2 \ge 4$.
