Nesbitt's inequality explained

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

	a
	b+c

	b
	a+c

	c
	a+b

\geq

	3
	2

with equality only when

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

(a+b),(b+c),(c+a)

	(a+b)+(a+c)+(b+c)
	3

\geq

\displaystyle

	1
	a+c

	1
	b+c

a+b

Clearing denominators yields

((a+b)+(a+c)+(b+c))\left(	1
	a+b

	1
	a+c

	1
	b+c

\right)\geq9,

from which we obtain

2	a+b+c
	b+c

	a+b+c
	a+c

	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

Supposing

a\geb\gec

, we have that

	1
	b+c

\ge

	1
	a+c

\ge

	1
	a+b

Define

\vec{x}=(a,b,c)

and

\vec{y}=\left(

	1
	b+c

	1
	a+c

	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

\vecy₁

and

\vecy₂

be the vector

\vecy

cyclically shifted by one and by two places; then

\vec{x} ⋅ \vec{y}\ge\vec{x} ⋅ \vecy₁

\vec{x} ⋅ \vec{y}\ge\vec{x} ⋅ \vecy₂

Addition then yields Nesbitt's inequality.

Third proof: Sum of Squares

The following identity is true for all

a,b,c:

	a
	b+c

	b
	a+c

	c
	a+b

	3
	2

	1	\left(
	2

	(a-b)²
	(a+c)(b+c)

	(a-c)²
	(a+b)(b+c)

	(b-c)²
	(a+b)(a+c)

\right).

This clearly proves that the left side is no less than

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	1
	\sqrt{a+b

},\frac,\frac\right\rangle yields

((b+c)+(a+c)+(a+b))\left(	1
	b+c

	1
	a+c

	1
	a+b

\right)\geq9,

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 obtain

	x+z
	y

	y+z
	x

	x+y
	z

\geq6,

because

	x
	y

	z
	y

	y
	x

	z
	x

	x
	z

	y
	z

\geq6\sqrt[6]{

	x
	y

⋅

	z
	y

⋅

	y
	x

⋅

	z
	x

⋅

	x
	z

⋅

	y
	z

} = 6.

Substituting out the

x,y,z

in favor of

a,b,c

yields

	2a+b+c
	b+c

	a+b+2c
	a+b

	a+2b+c
	c+a

\geq6

	2a
	b+c

	2c
	a+b

	2b
	a+c

+3\geq6,

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

real numbers

(x_k)

and any sequence of

positive numbers

(a_k)

	2
x
	k

a_k

\displaystyle\sum

\geq

k=1

(\sum

	2
x
	k)

k=1

	n
\sum		a_k
	k=1

We use the lemma on

(x_k)=(1,1,1)

and

(a_{k)=(b+c,a+c,a+b)}

. This gives

	1
	b+c

	1
	c+a

	1
	a+b

\geq

	3²
	2(a+b+c)

which results in

	a+b+c
	b+c

	a+b+c
	c+a

	a+b+c
	a+b

\geq

	9
	2

i.e.,

	a
	b+c

	b
	c+a

	c
	a+b

\geq

	9
	2

-3=

	3
	2

Seventh proof: Using homogeneity

As the left side of the inequality is homogeneous, we may assume

a+b+c=1

. Now define

x=a+b

y=b+c

, and

z=c+a

. The desired inequality turns into

	1-x
	x

	1-y
	y

	1-z
	z

\ge

	3
	2

, or, equivalently,

	1
	x

	1
	y

	1
	z

\ge

	9
	2

. This is clearly true by Titu's Lemma.

Eighth proof: Jensen's inequality

Let

S=a+b+c

and consider the function

f(x)=	x
	S-x

. This function can be shown to be convex in

[0,S]

and, invoking Jensen's inequality, we get

\displaystyle

	b
	S-b

	c
	S-c

S-a

\geq

	S/3
	S-S/3

A straightforward computation then yields

	a
	b+c

	b
	c+a

	c
	a+b

\geq

	3
	2

Ninth proof: Reduction to a two-variable inequality

By clearing denominators,

	a
	b+c

	b
	a+c

	c
	a+b

\geq

	3
	2

\iff2(a^3+b^3+c³⁾\geqab²+a^2b+ac²+a^2c+bc²+b^2c.

It therefore suffices to prove that

x^3+y³\geqxy^2+x^2y

for

(x,y)\in

	2
R
	+

, as summing this three times for

(x,y)=(a,b), (a,c),

and

(b,c)

completes the proof.

x^3+y³\geqxy^2+x^2y\iff(x-y)(x^2-y²⁾\geq0

we are done.

References

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

External links

See AoPS for more proofs of this inequality.