According to wikipedia, "let $f$ denote a real-valued, continuous and strictly increasing function on [0, c] with c > 0 and f(0) = 0. Let $f^{−1}$ denote the inverse function of $f$. Then, for all a ∈ [0, c] and b ∈ [0, f(c)],"
$$ab \le \int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx$$
I am wondering: does the inequality reverse if $f$ is decreasing, and $f(a) =0$ ? $($and, specifically with $a = b$, if that is necessary$)$
EDIT: I am pretty sure it neither holds nor reverses, as I can find examples of both. However, I don't understand why it doesn't hold. Consider $a=b=1$.
If I take an increasing function $f$, where $f(0) =0$ and $f(1) = 1$, everything is fine. But if I redefine $f$ as $f_{new}(x) = f(1-x)$ then the function is decreasing and the inequality falls apart.. even though the integral of $f_{new}$ equals that of the old, and I would think the same must be true for the inverses.
If $f$ is strictly decreasing on $[0,c]$, then $-f$ is strictly increasing on $[0,c]$. Then, noting that the inverse of $-f(x)$ is $f^{-1}(-x)$ we get that for all $a \in [0,c]$ and $b \in [0,-f(c)]$
$ab \leq -\int_0^a f(x) dx + \int_0^b f^{-1}(-x) dx$
(some of the intervals in the theorem doesn't make sense unless we agree that $(a,b)=(b,a)$ when $b < a$)