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?

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TopNewestIn 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 \(1\). – Jan J. · 4 years ago

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

(i am getting K as 1, which is possible wrong) – Sameer Arora · 2 years, 3 months ago

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– Cody Johnson · 2 years, 3 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)Log in to reply

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.

thank you in advance – Sameer Arora · 2 years, 3 months ago

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– Cody Johnson · 4 years ago

Do you have any other good examples of how homogeneity is useful?Log in to reply

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.) – Jan J. · 4 years ago

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@Jan J. – Surya Prakash · 1 year, 10 months ago

What should I assume, if the given inequality is homogenous equation of degree one?Log in to reply

– Cody Johnson · 4 years ago

Thank you so much!Log in to reply

– Shourya Pandey · 3 years, 12 months ago

A symmetric equation can be assumed true.Log in to reply

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=386799&p=2148037#p2148037 in here – Truong Nguyen Ngoc · 4 years ago

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