Find min $P$: $P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$ Let $a,b,c\geq 0$ and $a+b+c=1$. Know that never have two numbers both zero. Find min $P$:
$$P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$$

The question is tagged integral-inequality. Therefore, I assumed that an integer solution is sought for.
At least one of two integer variables must be unequal to zero
(= greater equal one):
$$a + b \geq 1$$
$$a + c \geq 1$$
$$b + c \geq 1$$
With $a+b+c = 1$ we get:
$$c = 1 - (a+b) \leq 0$$
Because $c \geq 0$, it follows that $c$ must be zero.
Therefore:
$$a \geq 1$$
$$b \geq 1$$
This contradicts $a+b+c=1$.
Thus, there is no integer solution fulfilling the constraint.
A non-integer solution would be $a = 0.5$ and $b = 0.5$.
This leads to $P = 8.0$.
