Find the smallest cosntant $k>0$ such that $\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \leq k(a+b+c)$ for every $a,b,c>0$. In the book 'Putnam and beyond', page $173$, it has the following problem: 

Find the smallest cosntant $k>0$ such that 
  $$\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b} \leq k(a+b+c)$$ for every $a,b,c>0$.

In the solution, it states that: 
Note that the inequality remains unchanged on replacing $a,b,c$ by $ta,tb,tc$ with $t>0$. Consequently, the smallest value of $k$ is the supremum of 
$$f(a,b,c) = \frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b}$$ over the domain $\Delta= \{ (a,b,c) :a,b,c>0, a+b+c=1 \}.$
Question: Why the smallest value of $k$ is the supremum of the function above on the domain? In particular, I do not understand why they only care about the domain and not other domain, and why the fact 'remain unchanged ...' will lead to the above conclusion. 

The inequality is equivalent to $f(a,b,c) \leq k$ where 
$$
f(a,b,c)=\frac{\frac{ab}{a+b+2c} + \frac{bc}{b+c+2a} + \frac{ca}{c+a+2b}}{a+b+c}
$$
Now, if we simplify $f(ta,tb,tc)$ we find that it is equal to $f(a,b,c)$ : it "remains
unchanged". For any $p_0=(a_0,b_0,c_0)$ with $a_0,b_0,c_0>0$, if we put $p_1=(a_1,b_1,c_1)=(t_1a_0,t_1b_0,t_1c_0)$ with $t_1=\frac{1}{a_0+b_0+c_0}$ then
$f(p_1)=f(p_0)$ and $p_1\in \Delta$. 
