Inequality Masterpiece

We have $a^{2}+b^{2}+c^{2}+(a+b+c)^{2}\leq4 \Leftrightarrow 2\geq a^{2}+b^{2}+c^{2}+ab+bc+ca$ (1)

We set $P=\frac{ab+1}{(a+b)^{2}}+\frac{bc+1}{(b+c)^{2}}+\frac{ca+1}{(c+a)^{2}} \Leftrightarrow2P=\frac{2ab+2}{(a+b)^{2}}+\frac{2bc+2}{(b+c)^{2}}+\frac{2ca+2}{(c+a)^{2}}$ (2)

From (1) and (2) we have $2P\geq\frac{2ab+(a^{2}+b^{2}+c^{2}+ab+bc+ca)}{(a+b)^{2}}+\frac{2bc+(a^{2}+b^{2}+c^{2}+ab+bc+ca)}{(b+c)^{2}}+\frac{2ca+(a^{2}+b^{2}+c^{2}+ab+bc+ca)}{(c+a)^{2}}$ $\Leftrightarrow 2P\geq3+\frac{(b+c)(c+a)}{(a+b)^{2}}+\frac{(c+a)(a+b)}{(b+c)^{2}}+\frac{(a+b)(b+c)}{(c+a)^{2}}\geq3+3=6 (AM-GM)$ $\Leftrightarrow 2P\geq6$
$\Leftrightarrow P\geq3$ .

Also the proofs of the inequality used are given below: https://brilliant.org/wiki/nesbitts-inequality/

Proof for the inequality used at the starting of the problem: