Nesbitt's inequality

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

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

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

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

Clearing denominators yields

$((a+b)+(a+c)+(b+c))(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c})\ge 9,$

from which we obtain

$2\frac{a+b+c}{b+c}+2\frac{a+b+c}{a+c}+2\frac{a+b+c}{a+b}\ge 9$

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

$\overrightarrow{x}=(a,b,c)$

$\overrightarrow{y}=(\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b})$

The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call ${\overrightarrow{y}}_1$ and ${\overrightarrow{y}}_2$ the vector $\overrightarrow{y}$ shifted by one and by two, we have:

$\overrightarrow{x}\cdot \overrightarrow{y}\ge \overrightarrow{x}\cdot {\overrightarrow{y}}_1$

$\overrightarrow{x}\cdot \overrightarrow{y}\ge \overrightarrow{x}\cdot {\overrightarrow{y}}_2$

Addition yields our desired Nesbitt's inequality.

Third proof: Sum of Squares

The following identity is true for all $a,b,c:$

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

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 $⟨\sqrt{a+b},\sqrt{b+c},\sqrt{c+a}⟩,⟨\frac{1}{\sqrt{a+b}},\frac{1}{\sqrt{b+c}},\frac{1}{\sqrt{c+a}}⟩$ yields

$((b+c)+(a+c)+(a+b))(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b})\ge 9,$

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 obtain the following

$\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\ge 6.$

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

Substituting out the $x,y,z$ in favor of $a,b,c$ yields

$\frac{2a+b+c}{b+c}+\frac{a+b+2c}{a+b}+\frac{a+2b+c}{c+a}\ge 6$

$\frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}+3\ge 6$

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 $n$ real numbers $(x_k)$ and any sequence of $n$ positive numbers $(a_k)$, $\sum _{k=1}^n\frac{x_k^2}{a_k}\ge \frac{(\sum _{k=1}^n{x}_k{)}^2}{\sum _{k=1}^n{a}_k}$.

We use the lemma on $(x_k)=(1,1,1)$ and $(a_k)=(b+c,a+c,a+b)$. This gives,

$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge \frac{3^2}{2(a+b+c)}$

This results in,

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

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{9}{2}-3=\frac{3}{2}$

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 $\frac{1-x}{x}+\frac{1-y}{y}+\frac{1-z}{z}\ge \frac{3}{2}$, or, equivalently, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge 9/2$. This is clearly true by Titu's Lemma.

Eighth proof: Jensen inequality

Define $S=a+b+c$ and consider the function $f(x)=\frac{x}{S-x}$. This function can be shown to be convex in $[0,S]$ and, invoking Jensen inequality, we get

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

A straightforward computation yields

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

Ninth proof: Reduction to a two-variable inequality

By clearing denominators,

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}⟺2(a^3+b^3+c^3)\ge a{b}^2+a^{2}b+a{c}^2+a^{2}c+b{c}^2+b^{2}c.$

It now suffices to prove that $x^3+y^3\ge x{y}^2+x^{2}y$ for $(x,y)\in {\mathbb{ℝ}}_{+}^2$, as summing this three times for $(x,y)=(a,b),(a,c),$ and $(b,c)$ completes the proof.

As $x^3+y^3\ge x{y}^2+x^{2}y⟺(x-y)(x^2-y^2)\ge 0$ we are done.

References

• Nesbitt, A. M. (1902). "Problem 15114". Educational Times 55. https://archive.org/details/educationaltimes55educ/page/232/mode/2up. 
• 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. http://www.mediafire.com/file/1mw1tkgozzu. 
• "Who was Alfred Nesbitt, the eponym of Nesbitt inequality". https://hsm.stackexchange.com/questions/14733/who-was-a-m-nesbitt-the-eponym-of-nesbitts-inequality. 

External links

• See AoPS for more proofs of this inequality.
• "Nesbitt's inequality". http://planetmath.org/?op=getobj&from=objects&id={{{id}}}. 
• "proof of Nesbitt's inequality". http://planetmath.org/?op=getobj&from=objects&id={{{id}}}.