Let $ ABC$ be a non-equilateral triangle. Denote by $ I$ the incenter and by $ O$ the circumcenter of the triangle $ ABC$. Prove that $ \angle AIO\leq\frac{\pi}{2}$ holds if and only if $ 2\cdot BC\leq AB+AC$.
It is only calculation
\[ \angle AIO\le\frac{\pi}{2}\Leftrightarrow\cos AIO\ge 0\Leftrightarrow IA^{2}+IO^{2}-OA^{2}\le 0\Leftrightarrow \frac{r^{2}}{\sin^{2}\frac{A}{2}}+R^{2}-2Rr-R^{2}\ge 0\Leftrightarrow \sin^{2}\frac{A}{2}\le \frac{r}{2R}\Leftrightarrow \frac{(s-b)(s-c)}{bc}\le \frac{2S^{2}}{sabc}=\frac{2s(s-a)(s-b)(s-c)}{sabc}\Leftrightarrow 2(s-a)\ge a\Leftrightarrow b+c\ge 2a \]
here I used $ OI^{2}=R^{2}-2Rr$ and $ IA=\frac{r}{\sin\frac{A}{2}}$
Some obvious observation: if we have triangle $ ABC$ with $ |AB|=|AC|$ and point $ D$ on side $ BC$ then $ BD \leq CD$ is equivalent to $ \angle BDA \geq 90^\circ$.
Let the $ M$ be the midpoint of arc $ BC$ not containing $ A$. Then $ A, I, M$ are collinear, the triangle $ AOM$ is isosceles and from the Ptolemy Theorem we have: $ |AM\parallel BC|=|MI|(|AB|+|AC|)$ (since it is well known that $ |MI|=|MB|=|MC|$). Now $ \angle AIO \leq 90^\circ$ is equivalent to $ |MI| \leq |AI|$ or $ |AM| \geq 2 |MI|$ or $ |MI|(|AB|+|AC|) = |AM\parallel BC| \geq 2|MI\parallel BC|$ or $ |AB|+|AC| \geq 2 |BC|$.