Basic Integral arithmetics I'm trying to reproduce a inequality I found which I don't really understand and I wonder if there's just something missing or my integration skills aren't good enough.
Problem
Define a function $f(t,x) := \int_0^x {\sigma(t,y)}dy $ for some bounded (from zero and infinity), differentiable function $\sigma$.
The inequality I found is
$$
(f(t,x)-f(0,x_0))^2\geq \bigg(\int_{x\wedge x_0}^{x\vee x_0} {\sigma(t,y)}dy\bigg)^2 + 2\int_{x\wedge x_0}^{x\vee x_0} {\sigma(t,y)}dy\int_0^{x\wedge x_0}\big({\sigma(t,y)}-{\sigma(0,y)}\big)dy,
$$
for $(x\wedge x_0)=\min(x,x_0)$ and $(x\vee x_0)=\max(x,x_0)$, but I can't find any ways to get to this result.
I'm grateful for any advice!
 A: I don't have a complete answer, but I've made some good progress in the case where $x \geq x_0$ so I figured I'd share.
Given $x \geq x_0$, we can rewrite the right-hand side of the inequality like this:
$$\bigg(\int_{x_0}^{x} {\sigma(t,y)}dy\bigg)^2 + 2\int_{x_0}^{x} {\sigma(t,y)}dy\int_0^{x_0}\big({\sigma(t,y)}-{\sigma(0,y)}\big)dy$$
Now based on some basic integral properties:
$$\int_{x_0}^{x} {\sigma(t,y)}dy = \int_{0}^{x} {\sigma(t,y)}dy - \int_{0}^{x_0} {\sigma(t,y)}dy = f(x,t) - f(x_0, t)$$
$$\int_0^{x_0}\big({\sigma(t,y)}-{\sigma(0,y)}\big)dy = \int_0^{x_0}{\sigma(t,y)}dy - \int_0^{x_0}{\sigma(0,y)}dy = f(x_0, t) - f(x_0, 0)$$
So now we can rewrite the right-hand side in terms of $f$ as:
$$\big(f(t,x) - f(t, x_0)\big)^2 + 2\big(f(t,x) - f(t, x_0)\big)\big(f(t, x_0) - f(0, x_0)\big) = $$
$$f^2(t, x) -f^2(t, x_0)-2f(t, x)f(0, x_0)+2f(t, x_0)f(0, x_0)$$
Now let's bring in the left-hand side of the inequality:
$$f^2(t,x) - 2f(t,x)f(0,x_0) + f^2(0,x_0) \geq f^2(t, x) -f^2(t, x_0)-2f(t, x)f(0, x_0)+2f(t, x_0)f(0, x_0)$$
$$\leftrightarrow f^2(0, x_0) \geq -f^2(t, x_0) + 2f(t, x_0)f(0, x_0)$$
$$\leftrightarrow f^2(0, x_0) - 2f(0, x_0)f(t, x_0) + f^2(t, x_0) \geq 0$$
$$\leftrightarrow (f(0, x_0) - f(t, x_0))^2 \geq 0$$
which is true for any real-valued $f$, which is a given if $\sigma$ is real-valued.
I've tried using the same logic on the $x < x_0$ case but so far I've been a bit stuck, possibly I'm missing something obvious but this is what I've got for now.
