Without-loss-of-generality question Going through solutions of IMO'09. Bumped into a without-loss-of-generality assumption that I can't comprehend.
Here's the statement of the problem:
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that
$$
\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}\leq\frac{3}{16}.
$$
Here's how they start in the solution:
We prove the homogenized inequality
$$
\frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
for all positive real numbers $a,b,c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a,b,c>0$, fulfilling this condition, the inequality
$$
\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\leq\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).
$$
And so on... If someone's interested in the rest, it's problem A2, solution 2 in this pdf: https://www.imo-official.org/problems/IMO2009SL.pdf
Why is there no loss of generality in such a choice of $a,b,c$?

Suppose that we know that if $a+b+c=1$ the homogenized inequality is true. Then if we have arbitrary $A$, $B$, $C$, with $A+B+C=x$, we can define $a=A/x$ and so on. When we plug this into the inequality, the $x$'s drop out, so that $A$, $B$, and $C$ also satisfy the inequality.
