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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 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 \(1\). Jan J. · 4 years ago

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@Jan J. 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) Sameer Arora · 2 years, 3 months ago

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@Sameer Arora 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) Cody Johnson · 2 years, 3 months ago

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@Cody Johnson 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.

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

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@Jan J. Do you have any other good examples of how homogeneity is useful? Cody Johnson · 4 years ago

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@Cody Johnson 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.) Jan J. · 4 years ago

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@Jan J. What should I assume, if the given inequality is homogenous equation of degree one? @Jan J. Surya Prakash · 1 year, 10 months ago

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@Jan J. Thank you so much! Cody Johnson · 4 years ago

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@Jan J. A symmetric equation can be assumed true. Shourya Pandey · 3 years, 12 months ago

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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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