proof of Nesbitt’s inequality

Starting from Nesbitt’s inequality

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

we transform the left hand side:

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

Now this can be transformed into:

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

Division by 3 and the right yields:

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

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

Title	proof of Nesbitt’s inequality
Canonical name	ProofOfNesbittsInequality
Date of creation	2013-03-22 12:37:01
Last modified on	2013-03-22 12:37:01
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	6
Author	mathwizard (128)
Entry type	Proof
Classification	msc 00A07