Why is the difference of the supremum and infimum of the absolute value of f(x) <= the difference of the supremum and infimum of f(x)?

Question

I am currently trying to prove the triangle inequality for Riemann Integrals:

which is that supposing f is Riemann integrable on [a,b], then |f| is riemann integrable on[a,b] and $ |\int_{b}^{a}f(x)dx| \leq \int_{a}^{b}|f(x)|dx$.

The latter part of the statement is very easy to prove, just consider the order property and that $ -|f| \leq f \leq |f|.$

I am having an issue on the first part , however:

I am taking the approach to try and prove:

U(|f|,P)-L(|f|,P)< U(f,P) - L(f,P) (where P is a partition on [a,b])

As then that implies |f| would be Riemann integrable by Riemann's criterion for integrability.

However, I have done some research on this, and I encountered this in my own working, and it requires me to prove that:

$\sup|f(x)|- \inf|f(x)| \leq \sup(f(x))-\inf(f(x))$ for x on [a,b]

I would be greatful if someone could give me a nudge in the right direction in how to start proving this (beyond just the inverse triangle inequality).

I believe I put the correct tags on this post, but if not , please tell me and I'll change them!

Edit: This question was already answered here, I apologise. The absolute value of a Riemann integrable function is Riemann integrable.