Find min $P$: $P=\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b)$

Question

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

Answer 1

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 $P = 8.0$.

Answer 2

\begin{align} P=&\frac{1}{(a+b)(b+c)}+\frac{1}{(c+a)(a+b)}+(c+1)(3+a+b) \\ =&\frac{1}{1-c}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)+(c+1)(4-c) \\ \ge&\frac{1}{1-c}\frac{4}{(b+c)+(c+a)}+(c+1)(4-c) \\ =&\frac{4}{1-c^2}+(c+1)(4-c)=:f(c), \end{align} the equality holds iff $a=b$. $$f'(c)=-2c+3+\frac{8c}{(c^2-1)^2}>0,\quad\textrm{ when }0\le c\le1.$$ So $\min f(c)=f(0)=8$. Hence, $\min P=\min f(c)=8$, when $a=b=1/2$ and $c=0$.