parmenides51 wrote: Let $ABC$ be a triangle with side-lengths $a, b, c$, inscribed in a circle with radius $R$ and let $I$ be ir's incenter. Let $P_1, P_2$ and $P_3$ be the areas of the triangles $ABI, BCI$ and $CAI$, respectively. Prove that $$\frac{R^4}{P_1^2}+\frac{R^4}{P_2^2}+\frac{R^4}{P_3^2}\ge 16$$ It's $$\sum_{cyc}\frac{1}{a^2}\geq\frac{(a+b+c)\prod\limits_{cyc}(a+b-c)^3}{a^4b^4c^4}$$and since by Schur $$abc\geq\prod_{cyc}(a+b-c),$$it's enough to prove that $$\sum_{cyc}a^2b^2\geq\sum_{cyc}(2a^2b^2-a^4)$$or $$\sum_{cyc}(a^2-b^2)^2\geq0.$$

One more solution can be found here: https://artofproblemsolving.com/community/q1h1863815p12615311

$R^4(\sum\dfrac{1}{P_1^2})\geq R^4(\sum \dfrac{1}{P_1})^2\cdot {1}{3}\geq R^4(\dfrac{9}{\sum P_1})^2\cdot \dfrac{1}{3}=R^4\cdot\dfrac{27}{(ABC)}$ So it suffice to show that $R^4\cdot\dfrac{27}{(ABC)}\geq 16\Leftrightarrow 27R^4\geq 16\cdot\dfrac{a^2b^2c^2}{16R^2}\Leftrightarrow 27R^6\geq a^2b^2c^2=2^6R^6\prod \sin^2 \angle A$ $\Leftrightarrow \dfrac{3\sqrt{3}}{8}\geq \prod \sin \angle A$ Let $f(x)=ln(\sin x),x\in (0,\pi)$.Then $f''(x)<0$ so by Jensen's inequality we have that $\sum ln(\sin \angle A)\geq 3ln(\sin \dfrac{\sum \angle A}{3})=3ln(\sin \dfrac{\pi}{3})=ln\dfrac{3\sqrt{3}}{8}\Leftrightarrow \dfrac{3\sqrt{3}}{8}\geq \prod \sin \angle A$