Mathematical inequality
In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers a, b and c,
![{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1bca096f74b2d3b063c74da9f1731b00fedfcd8)
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.
First proof: AM-HM inequality
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By the AM-HM inequality on
,
![{\displaystyle {\frac {(a+b)+(a+c)+(b+c)}{3}}\geq {\frac {3}{\displaystyle {\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d50d2381318c98aae99c6d04c997ab723a084706)
Clearing denominators yields
![{\displaystyle ((a+b)+(a+c)+(b+c))\left({\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}\right)\geq 9,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d78ca4c4d8af7c84bfa7fe05ad46af03ec1c3fe2)
from which we obtain
![{\displaystyle 2{\frac {a+b+c}{b+c}}+2{\frac {a+b+c}{a+c}}+2{\frac {a+b+c}{a+b}}\geq 9}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25028d1ec4085ff8b7689da1d8cf8d3bc220bbf5)
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
![{\displaystyle {\frac {1}{b+c}}\geq {\frac {1}{a+c}}\geq {\frac {1}{a+b}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d878c447b71da9f28063843ae4be09eb5f68a9dd)
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
![{\displaystyle {\vec {x}}\cdot {\vec {y}}\geq {\vec {x}}\cdot {\vec {y}}_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bc9bdb34f232abd401e43fae13bf8d97e1a76ca)
![{\displaystyle {\vec {x}}\cdot {\vec {y}}\geq {\vec {x}}\cdot {\vec {y}}_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76050040cc8890cadfb01bdfa96b6e4524e07948)
Addition then yields Nesbitt's inequality.
Third proof: Sum of Squares
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The following identity is true for all
![{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec8cddcaf162a1509f22053e879df555a0c28ab1)
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
![{\displaystyle ((b+c)+(a+c)+(a+b))\left({\frac {1}{b+c}}+{\frac {1}{a+c}}+{\frac {1}{a+b}}\right)\geq 9,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/180f0e25a4123633e20f64b95b0e92e88ab01c91)
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
![{\displaystyle {\frac {x+z}{y}}+{\frac {y+z}{x}}+{\frac {x+y}{z}}\geq 6,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b731cecb405b55949f1b2b77083bf9c94c16a94)
because
Substituting out the
in favor of
yields
![{\displaystyle {\frac {2a+b+c}{b+c}}+{\frac {a+b+2c}{a+b}}+{\frac {a+2b+c}{c+a}}\geq 6}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a35b8cedddc6de35017b415eb11427e15a7a0b2)
![{\displaystyle {\frac {2a}{b+c}}+{\frac {2c}{a+b}}+{\frac {2b}{a+c}}+3\geq 6,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60ec6d11031194890e55afdc148b6a1e644977a1)
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
![{\displaystyle {\frac {1}{b+c}}+{\frac {1}{c+a}}+{\frac {1}{a+b}}\geq {\frac {3^{2}}{2(a+b+c)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07009426f527c299cf7eb4441a6828123d28e51b)
which results in
i.e.,
![{\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {9}{2}}-3={\frac {3}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e3168a33669e0d760d481ea3b0e887324a41c36)
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
![{\displaystyle \displaystyle {\frac {{\frac {a}{S-a}}+{\frac {b}{S-b}}+{\frac {c}{S-c}}}{3}}\geq {\frac {S/3}{S-S/3}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f0fcf805cc7e5879ff3b9ae3ed41902ac615597)
A straightforward computation then yields
![{\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00ad3400e459831ddaaa20fd8291a121637a2270)
Ninth proof: Reduction to a two-variable inequality
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By clearing denominators,
![{\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}\iff 2(a^{3}+b^{3}+c^{3})\geq ab^{2}+a^{2}b+ac^{2}+a^{2}c+bc^{2}+b^{2}c.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fca19656a680d8c9952caea533c8ddf94423ff33)
It therefore suffices to prove that
for
, as summing this three times for
and
completes the proof.
As
we are done.