Help with verifying integral inequality. I am looking at problems from a released Fall 14 mock exam. The question in particular is number 2:

Let $f$ be a continuous function in $[0,1]$ satisfying the condition:
$$ \int_x^1 f(t) dt \geq \frac{1-x^2}{2}$$ for $x \in [0,1]$
Prove that:
$$\int_0^1 |f(x)|^2 dx \geq \int_0^1 xf(x)dx$$

This is what I have come up with so far:
First off we can "evaluate" the integral from $0$ to $1$:
$$\int_0^1 f(t) dt \geq \frac{1-0^2}{2} = \frac{1}{2}$$
Next we know from the Cauchy-Schwartz inequality:
$$\left|\int_0^1 f(x) dx\right|^2 \leq \int_0^1 |f(x)|^2 dx$$
So:
$$\int_0^1 |f(x)|^2 dx \geq \frac{1}{4}$$
Now for the other equation. I used integration by parts:
$$\int_0^1 xf(x)dx = xF(x) - \int_0^1 F(x) dx $$
$$\int_0^1 xf(x) dx \geq 1 \cdot \frac{1}{2} - \int_0^1 F(x) dx $$
But notice that:
$$F(x)|_0^1 \geq \frac{1}{2}$$
So:
$$\int_0^1 xf(x) dx \geq \frac{1}{2} - \frac{1}{2}$$
$$\int_0^1 xf(x) dx \geq 0$$
Which seems to be right so far (I could of course be wrong). I don't have enough info to close anything out, but it seems to be pointing in the right direction. Any further hints or corrections would be greatly appreciated.
 A: Define $g=f^{+}=\max(f,0)$, $h=f^-=-\min(f,0)$, we have $f=g-h$, $|f|=g+h$. $g\geq 0$, $h\geq 0$.
The given condition tells us $\int_0^1 (g-h-x) dx\geq 0\Rightarrow \int_0^1 g dx\geq \int_0^1 (h+x)dx$
For the inequality we need to prove, LHS-RHS=$$\int_0^1 ((g+h)^2-x(g-h))dx\\=\int_0^1 (g^2+2gh+h^2-xg+xh)dx\\\geq \int_0^1 (g(h+x)+2gh+h^2-xg+xh)dx\\=\int_0^1 (3gh+h^2+hx)dx\geq 0$$
A: Since $\frac{1-x^2}{2} = \int_x^1 t\, dt$, then the hypothesis implies $\int_x^1 [f(t) - t]\, dt \ge 0$ for all $x\in [0,1]$. Let $F(x) := \int_x^1 [f(t) - t]\, dt$, so that $F(x) \ge 0$ for all $x\in [0,1]$. We have
$$ \int_0^1 f(x)^2\, dx - \int_0^1 2xf(x)\, dx + \int_0^1 x^2\, dx = \int_0^1 [f(x) - x]^2\, dx \ge 0$$
Thus
$$ \int_0^1 f(x)^2\, dx - \int_0^1 xf(x)\, dx \ge \int_0^1 xf(x)\, dx - \int_0^1 x^2\, dx = \int_0^1 x[f(x) - x]\, dx$$
Using integration by parts and the fact that $F(x) \ge 0$ for all $x\in [0,1]$, we find
$$\int_0^1 x[f(x) - x]\, dx = \int_0^1 x(-F'(x))\, dx = -xF(x)\bigg|_0^1 +\int_0^1 F(x)\, dx = \int_0^1 F(x)\, dx \ge 0$$
Therefore
$$\int_0^1 f(x)^2\, dx - \int_0^1 xf(x)\, dx \ge 0$$
or
$$\int_0^1 \lvert f(x)\rvert^2\, dx \ge \int_0^1 xf(x)\, dx$$
