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
with equality when all the are equal.
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 the Cauchy-Schwarz Inequality, we have
as desired. Equality occurs when , i.e., when .
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 .
The AM-HM inequality for three variables,
is equivalent to
Setting yields the desired inequality.
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 or the fact that ,
Since , weighted AM-HM gives us
By Muirhead's and Cauchy
By Cauchy, since by Muirhead as
Another Interesting Method
Let And And Now, we get Also by AM-GM; and
By Muirhead's and expansion
Let . Expanding out we get that our inequality is equivalent to This means So it follows that we must prove So it follows that we must prove which immediately follows from Muirhead's.