Solution:
Let $r$ be the radius of the inscribed circle of the triangle $ABC$. We have that
$$P_{1}=\frac{r c}{2}, \quad P_{2}=\frac{r a}{2}, \quad P_{3}=\frac{r b}{2}$$
It follows that
$$\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}}=\frac{4}{r^{2}}\left(\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)$$
From Leibniz's relation we have that if $H$ is the orthocenter, then
$$OH^{2}=9 R^{2}-a^{2}-b^{2}-c^{2}$$
It follows that
$$9 R^{2} \geq a^{2}+b^{2}+c^{2}$$
Therefore, using the AM-HM inequality and then (1), we get
$$\frac{1}{c^{2}}+\frac{1}{a^{2}}+\frac{1}{b^{2}} \geq \frac{9}{a^{2}+b^{2}+c^{2}} \geq \frac{1}{R^{2}}$$
Finally, using Euler's inequality, namely that $R \geq 2 r$, we get
$$\frac{1}{P_{1}^{2}}+\frac{1}{P_{2}^{2}}+\frac{1}{P_{3}^{2}} \geq \frac{4}{r^{2} R^{2}} \geq \frac{16}{R^{4}}$$