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Inequality

HapaxOromenon3

How can I show that for all positive integers a, b, and c, a/b + b/c + c/a >= 3?

Reply 1

username1162972

Original post by HapaxOromenon3

How can I show that for all positive integers a, b, and c, a/b + b/c + c/a >= 3?

AM-GM inequality case n=3.

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Reply 2

HapaxOromenon3

Original post by physicsmaths

AM-GM inequality case n=3.

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Yes, that seems to work, with the list of integers being (ab^2, bc^2, ca^2). Thanks.

Reply 3

username1162972

Original post by HapaxOromenon3

Yes, that seems to work, with the list of integers being (ab^2, bc^2, ca^2). Thanks.

You can do it directly.
Multiplying these terms gives 1 so LHS/3>=1 hence he inequality.
Alternatively note that a->ka b->kb etc the inequality is not changed so homegenous, so you may assume something which gives a different type of solution albeit longer but more satisfying.

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Reply 4

username1162972

The homogenous stuff is just food for thought for other inequalities where such a fact is very useful and used to the absolute extreme.

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