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.

Proof

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First proof: AM-HM inequality

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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

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Supposing  , we have that

 

Define

  and  .

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.

Third proof: Sum of Squares

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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

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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

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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.

Sixth proof: Titu's lemma

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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

  i.e.,
 

Seventh proof: Using homogeneity

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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's inequality

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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

 

Ninth proof: Reduction to a two-variable inequality

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By clearing denominators,

 

It therefore suffices to prove that   for  , as summing this three times for   and   completes the proof.

As   we are done.

References

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  • Nesbitt, A. M. (1902). "Problem 15114". Educational Times. 55.
  • Ion Ionescu, Romanian Mathematical Gazette, Volume XXXII (September 15, 1926 - August 15, 1927), page 120
  • Arthur Lohwater (1982). "Introduction to Inequalities". Online e-book in PDF format.
  • "Who was Alfred Nesbitt, the eponym of Nesbitt inequality".
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