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

\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 $c=0$.
