In mathematics, Nesbitt's inequality states that for positive real numbers a, b and c,
\( \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}. \)
It is an elementary special case (N = 3) 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 (a+b),(b+c),(c+a), \)
\( \frac{(a+b)+(a+c)+(b+c)}{3}\geq\frac{3}{\displaystyle\frac{1}{a+b}+\frac{1}{a+c}+ \frac{1}{b+c}}. \)
Clearing denominators yields
\( ((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\geq 9, \)
from which we obtain
\( 2\frac{a+b+c}{b+c}+2\frac{a+b+c}{a+c}+2\frac{a+b+c}{a+b}\geq9 \)
by expanding the product and collecting like denominators. This then simplifies directly to the final result.
Second proof: Rearrangement
Suppose \( a \ge b \ge c \) , we have that
\( \frac 1 {b+c} \ge \frac 1 {a+c} \ge \frac 1 {a+b} \)
define
\( \vec x = (a, b, c) \)
\( \vec y = \left(\frac 1 {b+c} , \frac 1 {a+c} , \frac 1 {a+b}\right) \)
The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call \( \vec y_1 \) and \( \vec y_2 \) the vector \( \vec y \) shifted by one and by two, we have:
\(\vec x \cdot \vec y \ge \vec x \cdot \vec y_1 \)
\( \vec x \cdot \vec y \ge \vec x \cdot \vec y_2 \)
Addition yields Nesbitt's inequality.
Third proof: Sum of Squares
The following identity is true for all a a,b,c:
\( \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} = \frac{3}{2} + \frac{1}{2} \left(\frac{(a-b)^2}{(a+c)(b+c)}+\frac{(a-c)^2}{(a+b)(b+c)}+\frac{(b-c)^2}{(a+b)(a+c)}\right) \)
This clearly proves that the left side is no less than \( {\frac {3}{2}} \) 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 \( \displaystyle\left\langle\sqrt{a+b},\sqrt{b+c},\sqrt{c+a}\right\rangle,\left\langle\frac{1} {\sqrt{a+b}},\frac{1}{\sqrt{b+c}},\frac{1}{\sqrt{c+a}}\right\rangle \) yields
\( ((b+c)+(a+c)+(a+b))\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9, \)
which can be transformed into the final result as we did in the AM-HM proof.
Fifth proof: AM-GM
Let x=a+b,y=b+c,z=c+a. We then apply the AM-GM inequality to the set of six values \( \left\{x^2z,z^2x,y^2z,z^2y,x^2y,y^2x\right\} \) to obtain
\( \frac{\left(x^2z+z^2x\right)+\left(y^2z+z^2y\right)+\left(x^2y+y^2x\right)}{6}\geq\sqrt[6]{x^2z\cdot z^2x\cdot y^2z\cdot z^2y\cdot x^2y\cdot y^2x}=xyz. \)
Dividing by xyz/6 yields
\( \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq6. \)
Substituting out the x,y,z in favor of a,b,c yields
\( {\displaystyle {\frac {2a+b+c}{b+c}}+{\frac {a+b+2c}{a+b}}+{\frac {a+2b+c}{c+a}}\geq 6} \)
\( {\displaystyle {\frac {2a}{b+c}}+{\frac {2c}{a+b}}+{\frac {2b}{a+c}}+3\geq 6} \)
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 n {\displaystyle n} n real numbers (x_k) and any sequence of n positive numbers \( (a_{k}) \), \( \displaystyle\sum_{k=1}^n\frac{x_k^2}{a_k}\geq\frac{(\sum_{k=1}^n x_k)^2}{\sum_{k=1}^n a_k} \). We use its three-term instance with x {\displaystyle x} x-sequence a,b,c and a-sequence a(b+c),b(c+a),c(a+b):
\( \frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)} \geq\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)} \)
By multiplying out all the products on the lesser side and collecting like terms, we obtain
\( \frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)}\geq\frac{a^2+b^2+c^2+ 2(ab+bc+ca)}{2(ab+bc+ca)}, \)
which simplifies to
\( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{a^2+b^2+c^2}{2(ab+bc+ca)}+1. \)
By the rearrangement inequality, we have \( a^2+b^2+c^2\geq ab+bc+ca \) , so the fraction on the lesser side must be at least \( \displaystyle\frac{1}{2} \) . Thus,
\( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}.
Seventh proof: Homogeneous
As the left side of the inequality is homogeneous, we may assume a+b+c=1. Now define \( {\displaystyle x=a+b} \) , \( {\displaystyle y=b+c} \) , and \( {\displaystyle z=c+a} \) . The desired inequality turns into \( {\displaystyle {\frac {1-x}{x}}+{\frac {1-y}{y}}+{\frac {1-z}{z}}\geq {\frac {3}{2}}} \) , or, equivalently, \( {\displaystyle {\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}}\geq 9/2}. \) This is clearly true by Titu's Lemma.
Eighth proof: Jensen inequality
Define \( {\displaystyle S=a+b+c} \) and consider the function \( {\displaystyle f(x)={\frac {x}{S-x}}} \). This function can be shown to be convex in \({\displaystyle [0,S]} \) and, invoking Jensen inequality, we get
\( {\displaystyle \displaystyle {\frac {{\frac {a}{S-a}}+{\frac {b}{S-b}} +{\frac {c}{S-c}}}{3}}\geq {\frac {S/3}{S-S/3}}.}
\)
A straightforward computation yields
\( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}. \)
Ninth proof: Reduction to a two-variable inequality
By clearing denominators,
\( {\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}\iff 2(a^{3}+b^{3}+c^{3})\geq ab^{2}+a^{2}b+ac^{2}+a^{2}c+bc^{2}+b^{2}c.} \)
It now suffices to prove that \( {\displaystyle x^{3}+y^{3}\geq xy^{2}+x^{2}y} \) for \( {\displaystyle (x,y)\in \mathbb {R} _{+}^{2}} \), as summing this three times for \( {\displaystyle (x,y)=(a,b),\ (a,c),} \) and \( {\displaystyle (b,c)} \) completes the proof.
As \( {\displaystyle x^{3}+y^{3}\geq xy^{2}+x^{2}y\iff (x-y)(x^{2}-y^{2})\geq 0} \) we are done.
References
Nesbitt, A.M., Problem 15114, Educational Times, 3(2), 1903.
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.
Undergraduate Texts in Mathematics
Graduate Studies in Mathematics
Hellenica World - Scientific Library
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