Nesbitt's inequality


In mathematics, Nesbitt's inequality states that for positive real numbers a, b and c,
It is an elementary special case of the difficult and much studied Shapiro inequality, and was published at least 50 years earlier.
There is no corresponding upper bound as any of the 3 fractions in the inequality can be made arbitrarily large.

Proof

First proof: AM-HM inequality

By the AM-HM inequality on,
Clearing denominators yields
from which we obtain
by expanding the product and collecting like denominators. This then simplifies directly to the final result.

Second proof: Rearrangement

Suppose, we have that
define
The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call and the vector shifted by one and by two, we have:
Addition yields Nesbitt's inequality.

Third proof: Sum of Squares

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.

Fourth proof: Cauchy–Schwarz

Invoking the Cauchy–Schwarz inequality on the vectors yields
which can be transformed into the final result as we did in.

Fifth proof: AM-GM

Let. We then apply the AM-GM inequality to the set of six values to obtain
Dividing by yields
Substituting out the in favor of yields
which then simplifies to the final result.

Sixth proof: Titu's Screw lemma

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 its three-term instance with -sequence and -sequence :
By multiplying out all the products on the lesser side and collecting like terms, we obtain
which simplifies to
By the rearrangement inequality, we have, so the fraction on the lesser side must be at least. Thus,

Seventh proof: Homogeneous

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.

Eighth proof: Jensen inequality

Define and consider the function. This function can be shown to be convex in and, invoking Jensen inequality, we get
A straightforward computation yields

Ninth proof: Reduction to a two-variable inequality

By clearing denominators,
It now suffices to prove that for, as summing this three times for and completes the proof.
As we are done.