# Homogeneity in Inequalities

I was reading this proof of 2001 IMO Problem #2, proving for $$a,b,c\in\mathbb{R}^+$$ that $$\sum \frac{a}{\sqrt{a^2+8bc}}\ge1$$. In a proof using Jensen's inequality, it says

This inequality is homogeneous so we can assume without loss of generality $a+b+c=1$.

I've read multiple proofs saying we can assume whatever because the terms are homogeneous, but what can you assume without loss of generality from homogeneity?

Note by Cody Johnson
7 years, 1 month ago

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In this post I consider only inequalities in three variables, but it extends to any number of variables. Expression $f(a,b,c)$ is said to be homogeneous of degree $k$ if and only if there exists real $k$ such that for every $t > 0$ we have $t^k \cdot f(a,b,c) = f(ta,tb,tc)$ For instance in your example we have $f(a,b,c) = \sum_{\text{cyc}} \frac{a}{\sqrt{a^2 + 8bc}} - 1$ and this expression $f$ is homogeneous of degree $0$, i.e. $f(a,b,c) = f(ta,tb,tc)$

Okay, now finally why can the assumption $a + b + c = 1$ be made? Assume that $a + b + c = m$ for $m > 0$ i.e. $\frac{a}{m} + \frac{b}{m} + \frac{c}{m} = 1$ Let $a' = \frac{a}{m}$, $b' = \frac{b}{m}$, $c' = \frac{c}{m}$. Then $a' + b' + c' = 1$ But the homegeneity of degree $0$ tells us that $f(a',b',c') = f\left(\frac{a}{m} + \frac{b}{m} + \frac{c}{m}\right) = f(a,b,c)$ (in case it's not clear, we used $\frac{1}{m} = t$, remember that $t$ can be arbitrary positive real number). Hence proving $f(a',b',c') \geq 0$ is equivalent to proving $f(a,b,c) \geq 0$ and we have the nice condition that $a' + b' + c' = 1$

You can assume many other things (but only one assumption at a time), like $a = 1$ $b = 1$ $c = 1$ $abc = 1$ $ab + bc + ca = 1$ $a^2 + b^2 + c^2 = 1$ etc. also the number on right-hand side of these assumptions doesn't need to be $1$.

- 7 years, 1 month ago

Do you have any other good examples of how homogeneity is useful?

- 7 years, 1 month ago

In general, homogeneous inequalities are easier to work with, that's because most known inequalities (AM-GM, Cauchy-Schwarz, etc.) are themselves homogeneous, so they are easier to apply to homogeneous inequalities, that's not to say that you can't prove non-homogeneous inequalities with them, but it's just usually easier to prove homogeneous inequalities.

Another reason is that many techniques rely on homogeneity such as SHED or [PDF] sum of squares

EDIT: Deleted example because it was wrong. (I am too tired.)

- 7 years, 1 month ago

i got the above example but can u pls explain how it can be used in inequality like:

a^2/(a+1) + b^2/(b+1) >= 1/3, where a+b=1;

(i am getting K as 1, which is possible wrong)

- 5 years, 4 months ago

In this case you must homogenize (make it homogeneous) by taking $\frac{a^2}{a+1}=\frac{a^2}{(a+b)(2a+b)}$ and similar (hint: to finish, use Titus lemma)

- 5 years, 4 months ago

yes i did the same thing and solved the inequality. but i was trying to figure out K using the method described above. can you please tell the degree of homogeniety (K) in my problem.

i got K=1, but since this inequality is equivalent to : a^3 + b^3 >= ((a^2)b) + ((b^2)a), this gives K=3.

- 5 years, 4 months ago

Thank you so much!

- 7 years, 1 month ago

What should I assume, if the given inequality is homogenous equation of degree one? @Jan J.

- 4 years, 11 months ago

But how can we assume $a=1$?!

- 2 years, 5 months ago

A symmetric equation can be assumed true.

- 7 years, 1 month ago

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=386799&p=2148037#p2148037 in here

- 7 years, 1 month ago