Find all values of x so that arccos x > arcsin x + arctan x.
First, we know that $-1 \leq \sin x, \cos x \leq 1$, so $\frac{\pi}2 \geq \arcsin x, \arccos x \geq -\frac{\pi}2$ and $\frac{\pi}4 \geq \arctan x \geq -\frac{\pi}4$.
If $\arccos x = \arcsin x + \arctan x$,
\begin{align*}
\cos(\arccos x) &= \cos(\arcsin x + \arctan x) \\
x &= \cos(\arcsin x)\cos(\arctan x) - \sin(\arcsin x)\sin(\arctan x) \\
x &= \sqrt{1 - x^2} \cdot \frac{1}{\sqrt{1 + x^2}} - x \cdot \frac{x}{\sqrt{1 + x^2}}.
\end{align*}Squaring both sides (ignoring the extraneous solution $x = -1$),
\[x^2 = \frac{1 - x^2}{1 + x^2} - \frac{x^2}{1 + x^2} = \frac{1 - 2x^2}{1 + x^2}.\]Then $x^2 (1 + x^2) = 1 - 2x^2$, so $x^4 + 2x^2 - 1 = 0$. This factors as $(x^2 - 1)(x^2 + 1) = 0$, so $x = \boxed{-1, 1}$.
Note: We discarded $\cos(\arcsin x) = \sqrt{1 - x^2} < 0$ in our algebra, but this yields $\arccos x = \pi - \arcsin x$, which is also nonnegative (so our calculation still holds).