Difference between revisions of "Nesbitt's Inequality"
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Revision as of 13:27, 8 November 2007
Nesbitt's Inequality is a theorem which, although rarely cited, has many instructive proofs. It states that for positive ,
,
with equality when all the variables are equal.
All of the proofs below generalize to proof the following more general inequality.
If are positive and , then
,
or equivalently
,
with equality when all the are equal.
Contents
Proofs
By Rearrangement
Note that and , , are sorted in the same order. Then by the rearrangement inequality,
.
For equality to occur, since we changed to , we must have , so by symmetry, all the variables must be equal.
By Cauchy
By the Cauchy-Schwarz Inequality, we have
,
or
,
as desired. Equality occurs when , i.e., when .
We also present three closely related variations of this proof, which illustrate how AM-HM is related to AM-GM and Cauchy.
By AM-GM
By applying AM-GM twice, we have
,
which yields the desired inequality.
By Expansion and AM-GM
We consider the equivalent inequality
.
Setting , we expand the left side to obtain
,
which follows from , etc., by AM-GM, with equality when .
By AM-HM
The AM-HM inequality for three variables,
,
is equivalent to
.
Setting yields the desired inequality.
By Substitution
The numbers satisfy the condition . Thus it is sufficient to prove that if any numbers satisfy , then .
Suppose, on the contrary, that . We then have , and . Adding these inequalities yields , a contradiction.
By Normalization and AM-HM
We may normalize so that . It is then sufficient to prove
,
which follows from AM-HM.
By Weighted AM-HM
We may normalize so that .
We first note that by the rearrangement inequality,
,
so
.
Since , weighted AM-HM gives us
.