<s>[INST] 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. 
 [/INST] 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 $p_1\in \Delta$. 
</s>