# proof of Nesbitt’s inequality

Starting from Nesbitt’s inequality

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

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 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.

Title proof of Nesbitt’s inequality ProofOfNesbittsInequality 2013-03-22 12:37:01 2013-03-22 12:37:01 mathwizard (128) mathwizard (128) 6 mathwizard (128) Proof msc 00A07