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:

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.

Proof

First proof

Starting from Nesbitt's inequality(1903)

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}

we transform the left hand side:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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:

Failed to parse (Missing texvc executable; please see math/README to configure.): ((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:

Failed to parse (Missing texvc executable; please see math/README to configure.): \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.

Second proof

Suppose Failed to parse (Missing texvc executable; please see math/README to configure.): a \ge b \ge c , we have that

Failed to parse (Missing texvc executable; please see math/README to configure.): \frac 1 {b+c} \ge \frac 1 {a+c} \ge \frac 1 {a+b}

define

Failed to parse (Missing texvc executable; please see math/README to configure.): \vec x = (a, b, c)

Failed to parse (Missing texvc executable; please see math/README to configure.): \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 Failed to parse (Missing texvc executable; please see math/README to configure.): \vec y_1

and Failed to parse (Missing texvc executable; please see math/README to configure.): \vec y_2
the vector Failed to parse (Missing texvc executable; please see math/README to configure.): \vec y
shifted by one and by two, we have:

Failed to parse (Missing texvc executable; please see math/README to configure.): \vec x \cdot \vec y \ge \vec x \cdot \vec y_1

Failed to parse (Missing texvc executable; please see math/README to configure.): \vec x \cdot \vec y \ge \vec x \cdot \vec y_2

Addition yields Nesbitt's inequality.

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.fi:Nesbittin epäyhtälö vi:Bất đẳng thức Nesbitt