I have seen, in many proofs for inequalities, the author does something called normalization. I believe this is only possible for homogeneous inequalities. I saw this in a proof of Nesbitt's inequality:
$$\frac{a}{b+c} + \frac{b}{a +c} + \frac{c}{a+b} \geq \frac32$$
The above can be transformed as:
$$\frac{a+b+c}{b+c} + \frac{a+b+c}{a+c} + \frac{a+b+c}{a+b} - 3$$
$$=(a+b+c)\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3$$
$$=\frac{1}{2}[(a+b) + (b+c) + (c+a)]\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3$$
Now, the author says, since the inequality is homogeneous, we may normalize $a+b+c = 1$
From the $AM-HM$ inequality:
$$\frac{\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}}{3} \geq \frac{3}{2(a+b+c)}$$
$$\implies \frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b} \geq \frac92$$
Since $a+b+c = 1$, we may multiply the LHS with $\frac12(a+b+b+c+c+a)$, without changing its value
$$\implies \frac{1}{2}[(a+b) + (b+c) + (c+a)]\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) \geq \frac92$$
Subtracting $3$ from both sides gives us the result.
I don't understand the normalization procedure at all. I can't see why it is possible to do this. It also seems a bit too good to be true.
