Just use $1+\frac{r}{R} = \cos(a)+\cos(b)+\cos(c)$ and bash.

A cool problem! I will write a most geometric solution in the case of a difference of two angles equal to $90^{\circ}$ and a non-strict sign. Let the equality $\angle B - \angle C = 90^{\circ}$ be true in triangle $ABC$ with incenter and circumcenter $I,O$. Let $AD$ be a diameter in $(ABC)$ and let $R_{(ABC)}=1$. Then $ABCD$ is isosceles trapezoid (because $\angle DBC = \angle ACB = \angle B - 90^{\circ}$) and we need to prove that $OI^2 \geq \frac{1}{2}$ (then we are done by $OI^2=R^2-2Rr$). Let $S$ be a midpoint of arc $ABD$. Then $I \in AS$ and so $OI^2 \geq d(O, AS)^2 = \frac{1}{2}$, because $AO=OS=1$ and $\angle AOS = 90^{\circ}$. So, we are done!

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This can be done also by differentiating after trig

Let $r$ and $R$ denote the inradius and circumradius of $\triangle ABC$. It is known that $\frac{r}{R} = 4\sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}}$. WLOG say $A > 90^\circ + C \iff \frac{A-C}{2} > 45^\circ$ We need to prove that \[2\sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}} < \frac 18\]But this is true as \begin{align*} 2\sin{\frac{A}{2}} \sin{\frac{B}{2}} \sin{\frac{C}{2}} &= \left(\cos\frac{A-C}{2} - \cos\frac{A+C}{2}\right)\sin\frac B2\ \\ &< \left(\cos 45^\circ - \sin\frac B2\right)\sin\frac B2 \\ & = \frac 1{\sqrt2}\sin\frac B2 - \sin^2 \frac B2 \\ & = \frac 18 - \left(\frac 1{2\sqrt{2}} - \sin\frac B2\right)^2 \\ &\le \frac 18 \end{align*}

doing random stuff led to this clean solution [asy][asy] size(200); defaultpen(linewidth(0.6)+fontsize(10)); pair Ma = dir(10); pair Mb = dir(55); pair Mc = dir(125); pair D = dir(170); pair B = Ma+Mc-Mb; pair A = 2*Mc-B; pair C = 2*Ma-B; pair N9 = circumcenter(Ma,Mb,Mc); pair I = incenter(A,B,C); draw(incircle(A,B,C), palered); draw(arc(N9, circumradius(Ma,Mb,Mc), -10, 190), lightred); draw(A--B--C--A, blue); draw(A--D--B, blue+dotted); draw(D--Mc--Mb--Ma--Mc, lightblue); dot("$A$", A, dir(I--A)); dot("$B$", B, dir(I--B)); dot("$C$", C, dir(I--C)); dot("$M_A$", Ma, dir(-45)); dot("$M_B$", Mb, dir(N9--Mb)); dot("$M_C$", Mc, dir(170)); dot("$D$", D, dir(180)); dot("$I$", I, dir(10)); dot("$N_9$", N9); [/asy][/asy] Let the triangle be $\triangle{ABC}$ with medial triangle $\triangle{M_AM_BM_C}$, altitude ${\overline{AD}}$, nine-point center ${N_9}$, and incenter ${I}$. Suppose that $B-C > 90^{\circ}$. Because $M_AM_BM_CD$ is an isosceles trapezoid with circumcenter ${N_9}$, we have \[ \angle{M_AM_CD} = \angle{M_BM_CD} - \angle{M_BM_CM_A} = \angle{M_AM_BM_C} - \angle{BM_AM_C} = B-C > 90^{\circ} \]so ${A}$ and ${N_9}$ lie on opposite sides of $\overline{BC}$. In particular ${N_9}$ lies outside the incircle of $\triangle{ABC}$, so \[ \frac{R}{2} - r = IN_9 > r \implies R > 4r. \]$\blacksquare$