Nesbitt's inequality

In mathematics, Nesbitt's inequality states that for positive real a, b and c we have:

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

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.

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