Show that $\frac{P_a}{a^2}+\frac{P_b}{b^2}+\frac{P_c}{c^2}\ge\frac{3}{4R}$. When is the equality reached? Show that $\dfrac{P_a}{a^2}+\dfrac{P_b}{b^2}+\dfrac{P_c}{c^2}\ge\dfrac{3}{4R}$. When is the equality reached?
We're dealing with an acute triangle $ABC$ where $h_a,h_b$ and $h_c$ are altitudes. $P_a,P_b$ and $P_c$ are the distances from the vertices of the triangle to the segments joining the feet of the altitudes. $R$ is the radius of the circumscribed circle of $ABC$.

I discovered a few things in the given configuration. $$\dfrac{P_a}{h_a}=\cos\measuredangle BAC$$ For the proof, we can look at the similar triangles $ABC$ and $AB_1C_1$. In them $\dfrac{AB_1}{AB}=\dfrac{h_{B_1C_1}}{h_{BC}}=\dfrac{P_a}{h_a}=\cos\measuredangle BAC$ (using the definition of cosine of an acute angle in the right triangle $AB_1B$). To show that the mentioned triangles are similar we can use $\triangle AB_1B\sim\triangle AC_1C$ and then SAS.
Similarly, $$\dfrac{P_b}{h_b}=\cos\measuredangle ABC\\ \dfrac{P_c}{h_c}=\cos\measuredangle ACB.$$ Another thing I was able to prove is $$P_a+P_b+P_c=\dfrac{a^2+b^2+c^2}{4R}.$$ We can derive that by simply substituting $$P_a=h_a\cos\alpha,P_b=h_b\cos\beta,P_c=h_c\cos\gamma,h_a=\dfrac{2S}{a},h_b=\dfrac{2S}{b},h_c=\dfrac{2S}{c},$$ where $S$ is the area of triangle $ABC$ and $\alpha,\beta$ and $\gamma$ are the angles of the triangle, and the law of cosines for simplification. Thank you in advance!
 A: $$\dfrac{h_a}{P_a}=\cos(\alpha)$$
$$\dfrac{P_a}{a^2}+\dfrac{P_b}{b^2}+\dfrac{P_c}{c^2} \ge= 3\sqrt[3]{\dfrac {h_a h_b h_c}{a^2b^2c^2\cos(\alpha)\cos(\beta)\cos(\gamma)}}=\dfrac{3S}{2abc}\dfrac{1}{\sqrt[3]{\cos(\alpha)\cos(\beta)\cos(\gamma)}}=\dfrac{3}{8R}\dfrac{1}{\sqrt[3]{\cos(\alpha)\cos(\beta)\cos(\gamma)}}\ge \dfrac{3}{4R}$$
$$ 3\sqrt[3]{\cos(\alpha)\cos(\beta)\cos(\gamma)} \le  \cos(\alpha)+\cos(\beta)+\cos(\gamma)=1+\dfrac{r}{R}\le\dfrac{3}{2}$$
A: As you mentioned, $\triangle AB_1C_1 \sim \triangle ABC \ $.
So, $AC_1 = \cfrac{P_a}{\sin C} = \cfrac{2R \cdot P_a}{c}$
Similarly show that $\triangle A_1BC_1 \sim \triangle ABC \ $, $C_1B = \cfrac{2R \cdot P_b}{c}$
$c = AC_1 + C_1 B = \cfrac{2R (p_a + P_b)}{c} \implies P_a + P_b = \cfrac{c^2}{2R}$
Also, $P_b + P_c = \cfrac{a^2}{2R}, P_c + P_a = \cfrac{b^2}{2R}$
So we get, $P_a = \cfrac{b^2+c^2-a^2}{4R}$
$\sum_{cyc} \cfrac{P_a}{a^2} = \cfrac{1}{4R} \sum_{cyc} \cfrac{b^2+c^2-a^2}{a^2}$
$ = \cfrac{1}{4R} \sum_{cyc} \left (\cfrac{a^2}{b^2} + \cfrac{b^2}{a^2} - 1 \right)$
$\geq \cfrac{3}{4R} \ $ (by AM-GM)
A: By your work we need to prove that:
$$\sum_{cyc}\frac{S\cos\alpha}{4R^3\sin^3\alpha}\geq\frac{3}{4R}$$ or
$$\sum_{cyc}\frac{\cos\alpha}{\sin^3\alpha}\geq\frac{3R^2}{S}$$ or
$$\sum_{cyc}\frac{b^2+c^2-a^2}{2bc\cdot\frac{8S^3}{b^3c^3}}\geq\frac{3a^2b^2c^2}{16S^3}$$ or
$$\sum_{cyc}b^2c^2(b^2+c^2-a^2)\geq3a^2b^2c^2$$ or
$$\sum_{cyc}c^2(a^2-b^2)^2\geq0.$$
