In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers a, b and c,
with equality only when (i. e. in an equilateral triangle).
There is no corresponding upper bound as any of the 3 fractions in the inequality can be made arbitrarily large.
It is the three-variable case of the rather more difficult Shapiro inequality, and was published at least 50 years earlier.
Clearing denominators yields
from which we obtain
by expanding the product and collecting like denominators. This then simplifies directly to the final result.
Supposing , we have that
Define
By the rearrangement inequality, the dot product of the two sequences is maximized when the terms are arranged to be both increasing or both decreasing. The order here is both decreasing. Let and be the vector cyclically shifted by one and by two places; then
Addition then yields Nesbitt's inequality.
The following identity is true for all
This clearly proves that the left side is no less than for positive a, b and c.
Note: every rational inequality can be demonstrated by transforming it to the appropriate sum-of-squares identity—see Hilbert's seventeenth problem.
Invoking the Cauchy–Schwarz inequality on the vectors yields
which can be transformed into the final result as we did in the AM-HM proof.
Let . We then apply the AM-GM inequality to obtain
because
Substituting out the in favor of yields
which then simplifies to the final result.
Titu's lemma, a direct consequence of the Cauchy–Schwarz inequality, states that for any sequence of real numbers and any sequence of positive numbers ,
We use the lemma on and . This gives
which results in
As the left side of the inequality is homogeneous, we may assume . Now define , , and . The desired inequality turns into , or, equivalently, . This is clearly true by Titu's Lemma.
Let and consider the function . This function can be shown to be convex in and, invoking Jensen's inequality, we get
A straightforward computation then yields
By clearing denominators,
It therefore suffices to prove that for , as summing this three times for and completes the proof.
As we are done.
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