Nesbitt's inequality

From Wikipedia, the free encyclopedia

You have new messages (last change).

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}.

Contents

[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

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.

Retrieved from "http://en.wikipedia.org../../../n/e/s/Nesbitt%27s_inequality.html"
In other languages