Let $f:[a,b]\to\mathbb{R}$ be a bounded function and set $M=\sup_{[a,b]}f(x)\,,\; m=\inf_{[a,b]}f(x)\,,\;M^*=\sup_{[a,b]}|f(x)|\,,\;m^*=\inf_{[a,b]}|f(x)|\,.$ Prove that $M^*-m^*\le M-m$.
First, since $f$ is bounded, all of $M$, $M^*$, $m$, $m^*$ are finite.
Next, we have $M^*-m^*\le M-m$ if and only if $0\le M-m+m^*-M^*=(M-M^*)+(m^*-m)\,,$ so it suffices to show that $M-M^*\ge 0$ and $m^*-m\ge 0$.
We obviously have $f\le |f|$, so it follows that $\inf f\le \inf|f|$; that is, $m\le m^*$. So $0\le m^*-m$.
But I can't figure out the other inequality. Thanks!