Nesbitt's inequality

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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:

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

1 Proof
2 Note

[edit] Proof

[edit] First proof

Starting from Nesbitt's inequality(1903)

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

we transform the left hand side:

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

Now this can be transformed into:

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

Division by 3 and the right factor yields:

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

Now on the left we have the arithmetic mean and on the right the harmonic mean, so this inequality is true.

We might also want to try to use GM for three variables.

[edit] Second proof

Suppose $a \ge b \ge c$ , we have that

$\frac 1 {b+c} \ge \frac 1 {a+c} \ge \frac 1 {a+b}$

define

$\vec x = (a, b, c)$

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

The scalar product of the two sequences is maximum because of the Rearrangement inequality if they are arranged the same way, call $\vec y_1$ and $\vec y_2$ the vector $\vec y$ shifted by one and by two, we have:

$\vec x \cdot \vec y \ge \vec x \cdot \vec y_1$

$\vec x \cdot \vec y \ge \vec x \cdot \vec y_2$

[edit] Third proof (using the identity).

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 then $\frac{3}{2}$ 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.

See this for more proofs of this inequality.

Addition yields Nesbitt's inequality.

[edit] Note

This article incorporates material from Nesbitt's inequality on PlanetMath, which is licensed under the GFDL. This article incorporates material from proof of Nesbitt's inequality on PlanetMath, which is licensed under the GFDL.

Categories: Inequalities

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Nesbitt's inequality

From Wikipedia, the free encyclopedia

Contents

[edit] Proof

[edit] First proof

[edit] Second proof

[edit] Third proof (using the identity).

[edit] Note

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