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.