Let, the sides of the triangle $ABC$ are $a, b, c$ units and the corresponding heights are $h_1, h_2, h_3$ units respectively.

Then the circumference is $a+b+c=2S$ and area is $P=\frac{1}{2}\left(a{h}_1\right)=\frac{1}{2}\left(b{h}_2\right)=\frac{1}{2}\left(c{h}_3\right)$

$\Rightarrow h_1=\frac{2P}{a}, h_2=\frac{2P}{b}, h_3=\frac{2P}{c}.$

Now $P^2\left[\frac{{\left(h_1{h}_2+h_2{h}_3+h_3{h}_1\right)}^2}{h_1^2{h}_2^2{h}_3^2}-2\right]$

$=P^2\left[\frac{{\left({\displaystyle \frac{2P}{a}}\times {\displaystyle \frac{2P}{b}}+{\displaystyle \frac{2P}{b}}\times {\displaystyle \frac{2P}{c}}+{\displaystyle \frac{2P}{c}}\times {\displaystyle \frac{2P}{a}}\right)}^2}{{\left({\displaystyle \frac{2P}{a}}\right)}^2{\left({\displaystyle \frac{2P}{b}}\right)}^2{\left({\displaystyle \frac{2P}{c}}\right)}^2}-2\right]$

$=P^2\left[\frac{16P^4{\left({\displaystyle \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\right)}^2}{\left({\displaystyle \frac{64P^6}{a^2{b}^2{c}^2}}\right)}-2\right]$

$=P^2\left[\frac{{\displaystyle \frac{{\left(a+b+c\right)}^2}{a^2{b}^2{c}^2}}}{\left({\displaystyle \frac{4P^2}{a^2{b}^2{c}^2}}\right)}-2\right]$

$=P^2\left[\frac{{\left(2S\right)}^2}{4P^2}-2\right]$

$=S^2-2P^2.$