00A07-ProofOfNesbittsInequality.tex

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\pmtitle{proof of Nesbitt's inequality}
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\begin{document}
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 \PMlinkescapetext{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.
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