Nesbitt's inequality

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:

${\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}$

Proof

First proof: AM-HM inequality

By the AM-HM inequality on ${\displaystyle (a+b),(b+c),(c+a)}$,

${\displaystyle {\frac {(a+b)+(a+c)+(b+c)}{3}}\geq {\frac {3}{\displaystyle {\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}}}.}$

Clearing denominators yields

${\displaystyle ((a+b)+(a+c)+(b+c))\left({\frac {1}{a+b}}+{\frac {1}{a+c}}+{\frac {1}{b+c}}\right)\geq 9,}$

from which we obtain

${\displaystyle 2{\frac {a+b+c}{b+c}}+2{\frac {a+b+c}{a+c}}+2{\frac {a+b+c}{a+b}}\geq 9}$

by expanding the product and collecting like denominators. This then simplifies directly to the final result.

Second proof: Rearrangement

Suppose ${\displaystyle a\geq b\geq c}$, we have that

${\displaystyle {\frac {1}{b+c}}\geq {\frac {1}{a+c}}\geq {\frac {1}{a+b}}}$

define

${\displaystyle {\vec {x}}=(a,b,c)}$
${\displaystyle {\vec {y}}=\left({\frac {1}{b+c}},{\frac {1}{a+c}},{\frac {1}{a+b}}\right)}$

The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call ${\displaystyle {\vec {y}}_{1}}$ and ${\displaystyle {\vec {y}}_{2}}$ the vector ${\displaystyle {\vec {y}}}$ shifted by one and by two, we have:

${\displaystyle {\vec {x}}\cdot {\vec {y}}\geq {\vec {x}}\cdot {\vec {y}}_{1}}$
${\displaystyle {\vec {x}}\cdot {\vec {y}}\geq {\vec {x}}\cdot {\vec {y}}_{2}}$

Third proof: Hilbert's Seventeenth Problem

The following identity is true for all ${\displaystyle a,b,c:}$

${\displaystyle {\frac {a}{b+c}}+{\frac {b}{a+c}}+{\frac {c}{a+b}}={\frac {3}{2}}+{\frac {1}{2}}\left({\frac {(a-b)^{2}}{(a+c)(b+c)}}+{\frac {(a-c)^{2}}{(a+b)(b+c)}}+{\frac {(b-c)^{2}}{(a+b)(a+c)}}\right)}$

This clearly proves that the left side is no less than ${\displaystyle {\frac {3}{2}}}$ 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

Invoking the Cauchy–Schwarz inequality on the vectors ${\displaystyle \displaystyle \left\langle {\sqrt {a+b}},{\sqrt {b+c}},{\sqrt {c+a}}\right\rangle ,\left\langle {\frac {1}{\sqrt {a+b}}},{\frac {1}{\sqrt {b+c}}},{\frac {1}{\sqrt {c+a}}}\right\rangle }$ yields

${\displaystyle ((b+c)+(a+c)+(a+b))\left({\frac {1}{b+c}}+{\frac {1}{a+c}}+{\frac {1}{a+b}}\right)\geq 9,}$

which can be transformed into the final result as we did in the AM-HM proof.

Fifth proof: AM-GM

We first employ a Ravi substitution: let ${\displaystyle x=a+b,y=b+c,z=c+a}$. We then apply the AM-GM inequality to the set of six values ${\displaystyle \left\{x^{2}z,z^{2}x,y^{2}z,z^{2}y,x^{2}y,y^{2}x\right\}}$ to obtain

${\displaystyle {\frac {\left(x^{2}z+z^{2}x\right)+\left(y^{2}z+z^{2}y\right)+\left(x^{2}y+y^{2}x\right)}{6}}\geq {\sqrt[{6}]{x^{2}z\cdot z^{2}x\cdot y^{2}z\cdot z^{2}y\cdot x^{2}y\cdot y^{2}x}}=xyz.}$

Dividing by ${\displaystyle xyz/6}$ yields

${\displaystyle {\frac {x+z}{y}}+{\frac {y+z}{x}}+{\frac {x+y}{z}}\geq 6.}$

Substituting out the ${\displaystyle x,y,z}$ in favor of ${\displaystyle a,b,c}$ 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 ${\displaystyle n}$ real numbers ${\displaystyle (x_{k})}$ and any sequence of ${\displaystyle n}$ positive numbers ${\displaystyle (a_{k})}$, ${\displaystyle \displaystyle \sum _{k=1}^{n}{\frac {x_{k}^{2}}{a_{k}}}\geq {\frac {(\sum _{k=1}^{n}x_{k})^{2}}{\sum _{k=1}^{n}a_{k}}}}$. We use its three-term instance with ${\displaystyle x}$-sequence ${\displaystyle a,b,c}$ and ${\displaystyle a}$-sequence ${\displaystyle a(b+c),b(c+a),c(a+b)}$:

${\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {(a+b+c)^{2}}{a(b+c)+b(c+a)+c(a+b)}}}$

By multiplying out all the products on the lesser side and collecting like terms, we obtain

${\displaystyle {\frac {a^{2}}{a(b+c)}}+{\frac {b^{2}}{b(c+a)}}+{\frac {c^{2}}{c(a+b)}}\geq {\frac {a^{2}+b^{2}+c^{2}+2(ab+bc+ca)}{2(ab+bc+ca)}},}$

which simplifies to

${\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {a^{2}+b^{2}+c^{2}}{2(ab+bc+ca)}}+1.}$

By the rearrangement inequality, we have ${\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ca}$, so the fraction on the lesser side must be at least ${\displaystyle \displaystyle {\frac {1}{2}}}$. Thus,

${\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}$

Seventh proof: Homogeneous

As the left side of the inequality is homogeneous, we may assume ${\displaystyle a+b+c=1}$. 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 funcion 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

${\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {3}{2}}.}$

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.