Nesbitt's inequality

In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers a, b and c,

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

with equality only when ${\displaystyle a=b=c}$ (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

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

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

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

Define

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

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 ${\displaystyle {\vec {y}}_{1}}$ and ${\displaystyle {\vec {y}}_{2}}$ be the vector ${\displaystyle {\vec {y}}}$ cyclically shifted by one and by two places; then

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

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

Addition then yields Nesbitt's inequality.

Third proof: Sum of Squares

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 ${\displaystyle 3/2}$ 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 ${\displaystyle \displaystyle \left\langle {\sqrt {a+b}},{\sqrt {b+c}},{\sqrt {c+a}}\right\rangle ,\left\langle {\frac {1}{\sqrt {a+b}}},{\frac {1}{\sqrt {b+c}}},{\frac {1}{\sqrt {c+a}}}\right\rangle }$ 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

Let ${\displaystyle x=a+b,y=b+c,z=c+a}$. We then apply the AM-GM inequality to obtain

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

because ${\displaystyle {\frac {x}{y}}+{\frac {z}{y}}+{\frac {y}{x}}+{\frac {z}{x}}+{\frac {x}{z}}+{\frac {y}{z}}\geq 6{\sqrt[{6}]{{\frac {x}{y}}\cdot {\frac {z}{y}}\cdot {\frac {y}{x}}\cdot {\frac {z}{x}}\cdot {\frac {x}{z}}\cdot {\frac {y}{z}}}}=6.}$

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 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 the lemma on ${\displaystyle (x_{k})=(1,1,1)}$ and ${\displaystyle (a_{k})=(b+c,a+c,a+b)}$. This gives

${\displaystyle {\frac {1}{b+c}}+{\frac {1}{c+a}}+{\frac {1}{a+b}}\geq {\frac {3^{2}}{2(a+b+c)}},}$

which results in

${\displaystyle {\frac {a+b+c}{b+c}}+{\frac {a+b+c}{c+a}}+{\frac {a+b+c}{a+b}}\geq {\frac {9}{2}}}$ i.e.,

${\displaystyle {\frac {a}{b+c}}+{\frac {b}{c+a}}+{\frac {c}{a+b}}\geq {\frac {9}{2}}-3={\frac {3}{2}}.}$

Seventh proof: Using homogeneity

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, ${\displaystyle {\frac {1}{x}}+{\frac {1}{y}}+{\frac {1}{z}}\geq {\frac {9}{2}}}$. This is clearly true by Titu's Lemma.

Eighth proof: Jensen's inequality

Let ${\displaystyle S=a+b+c}$ 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's inequality, we get

${\displaystyle \displaystyle {\frac {{\frac {a}{S-a}}+{\frac {b}{S-b}}+{\frac {c}{S-c}}}{3}}\geq {\frac {S/3}{S-S/3}}.}$

A straightforward computation then yields

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

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 therefore 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. (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".

External links

• See AoPS for more proofs of this inequality.
• "Nesbitt's inequality". PlanetMath .
• "proof of Nesbitt's inequality". PlanetMath .