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
4 years, 8 months 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 $$1$$.

- 4 years, 8 months 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)

- 2 years, 11 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)

- 2 years, 11 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.

- 2 years, 11 months ago

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

- 4 years, 8 months 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.)

- 4 years, 8 months ago

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

- 3 days, 14 hours ago

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

- 2 years, 6 months ago

Thank you so much!

- 4 years, 8 months ago

A symmetric equation can be assumed true.

- 4 years, 8 months ago

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

- 4 years, 8 months ago