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:

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

Proof

First proof: AM-HM

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: Hilbert's Seventeenth Problem

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

We first employ a Ravi substitution: 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

\frac{2a+b+c}{b+c}+\frac{a+b+2c}{a+b}+\frac{a+2b+c}{c+a}\geq6,

which then simplifies directly 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 $(a_k)$ , $\displaystyle\sum_{k=1}^n\frac{x_k^2}{a_k}\geq\frac{(\sum_{k=1}^n x_k)^2}{\sum_{k=1}^n a_k}$ . We use its three-term instance with $x$ -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}.

References

Arthur Lohwater (1982). "Introduction to Inequalities". Online e-book in PDF format.

External links

See AoPS for more proofs of this inequality.
Nesbitt's inequality at PlanetMath.org.
proof of Nesbitt's inequality at PlanetMath.org.

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