Without-loss-of-generality question

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 $a,b,c$?

Because the inequality is homogeneous. – Angina Seng May 21 '18 at 17:47 — Angina Seng, May 21 '18 at 17:47

score 2 · Accepted Answer · answered May 21 '18 at 17:50

There is no loss of generality because the inequality

$$ \frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) $$ is homogenized. This means: if you multiply $a,b$ and $c$ with the same constant $r$, the inequality does not change. Now give yourself some $a,b$ and $c$ where you have free choice of selecting these variables. You will have the sum $s = a + b+ c$. Now multiply $a,b$ and $c$ with the same constant $r=1/s$, then in the new (multiplied) variables you have $1 = a + b+ c$ which means you can always take that choice.

... where the problem statement excludes the possibility that $s=0$ — Hagen von Eitzen, May 21 '18 at 18:01

score 1 · Answer 2 · answered May 21 '18 at 17:53

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.

Without-loss-of-generality question

2 Answers2