Prove (or derive) $[X|X≤x]⩽x<[X|X>x]$

Question

How can I prove (or derive) the following law on conditional expectation: $$[X|X≤x]⩽x<[X|X>x]$$ It appears in this post but I don't know where this identity comes from.

Answer 1

HINT LHS says $$ \int_{-\infty}^x sf(s)ds \le x\int_{-\infty}^x f(s) ds $$ and in the LHS integral, $s <x$ because you are integrating over $(-\infty,x)$...

Can you finish this and do the RHS the same way?

HINT 2

Note that $$ \int_{-\infty}^x sf(s)ds \le \int_{-\infty}^x xf(s)ds = x\int_{-\infty}^x f(s) ds $$