Khang's inequality

We have:

$P=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}-\dfrac{1}{\sqrt[5]{ab+bc+ca}}$

$\quad=\dfrac{(a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1)}{(a+1)(b+1)(c+1)}- \dfrac{1}{\sqrt[5]{ab+bc+ca}}$

$\quad=\dfrac{ab+bc+ca+2(a+b+c)+3}{abc+ab+bc+ca+a+b+c+1}-\dfrac{1}{\sqrt[5]{ab+bc+ca}}$

$\quad\le\dfrac{ab+bc+ca+9}{ab+bc+ca+4}-\dfrac{1}{\sqrt[5]{ab+bc+ca}}$

$\quad=1+\dfrac{5}{ab+bc+ca+4}-\dfrac{1}{\sqrt[5]{ab+bc+ca}}$

$\quad\le1+\dfrac{5}{5\sqrt[5]{ab+bc+ca}}-\dfrac{1}{\sqrt[5]{ab+bc+ca}}=1$ (using AM-GM inequality)

The equality holds if and only if $(a, b, c)=(0, \dfrac{3+\sqrt5}{2}, \dfrac{3-\sqrt5}{2})$ or any permutation.

So, the maximum value of $P$ is $\boxed{1}$ .