In mathematics, Nesbitt's inequality is a special case of the Shapiro inequality. It states that for positive real numbers a, b and c we have:
Contents
- First proof AM HM
- Second proof Rearrangement
- Third proof Hilberts Seventeenth Problem
- Fourth proof CauchySchwarz
- Fifth proof AM GM
- Sixth proof Titus lemma
- References
First proof: AM-HM
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
define
The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call
Addition yields Nesbitt's inequality.
Third proof: Hilbert's Seventeenth Problem
The following identity is true for all
This clearly proves that the left side is no less than
Note: every rational inequality can be solved by transforming it to the appropriate identity, see Hilbert's seventeenth problem.
Fourth proof: Cauchy–Schwarz
Invoking the Cauchy–Schwarz inequality on the vectors
which can be transformed into the final result as we did in the AM-HM proof.
Fifth proof: AM-GM
We first employ a Ravi substitution: let
Dividing by
Substituting out the
which then simplifies directly to the final result.
Sixth proof: Titu's lemma
Titu's lemma, a direct consequence of the Cauchy–Schwarz inequality, states that for any sequence of
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