Nesbitt's inequality

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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:

Proof[edit]

First proof: AM-HM[edit]

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[edit]

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: Hilbert's Seventeenth Problem[edit]

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 solved by transforming it to the appropriate identity, see Hilbert's seventeenth problem.

Fourth proof: Cauchy–Schwarz[edit]

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.

Fifth proof: AM-GM[edit]

We first employ a Ravi substitution: 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 directly to the final result.

Sixth proof: Titu's lemma[edit]

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,

References[edit]

External links[edit]