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 $ABC$.

[![enter image description here](https://i.stack.imgur.com/EEqHc.png)](https://i.stack.imgur.com/EEqHc.png)

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!