Ahiles wrote: Determine the maximum possible real value of the number $ k$, such that \[ (a + b + c)(\frac {1}{a + b} + \frac {1}{c + b} + \frac {1}{a + c} - k )\ge k \] for all real numbers $ a,b,c\ge 0$ with $ a + b + c = ab + bc + ca$ $ (a + b + c)\left(\frac {1}{a + b} + \frac {1}{c + b} + \frac {1}{a + c} - k \right)\ge k\Leftrightarrow\frac{(ab+ac+bc)(a+b+c)}{(a+b)(a+c)(b+c)}\geq k$ Thus, the answer is $ 1.$

let $ a=0,b=c=2$ so $ k \leq 1$. $ a+b+c=ab+bc+ca$ so $ 1=\frac {ab+bc+ca}{a+b+c}$ $ \sum a .\sum \frac {1}{a+b} \geq k.(\sum a + \frac {\sum ab}{\sum a})$. so $ \sum ab.\frac {\sum a^2+3\sum ab}{\prod (a+b)} \geq k.\frac {\sum a^2+3\sum ab}{\sum a}$ so $ \sum a.\sum ab \geq k.\prod (a+b)$ let k=1 so that $ abc \geq 0$. when $ a=0$ or $ b=0$ or $ c=0$ $ \sum a .(\sum \frac {1}{a+b} -1)=1$