Nesbitt's Inequality
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Theorem
Let $a$, $b$ and $c$ be positive real numbers.
Then:
- $\dfrac {a} {b+c} + \dfrac {b} {a+c} + \dfrac {c} {a+b} \ge \dfrac 3 2$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \frac {a} {b+c} + \frac {b} {a+c} + \frac {c} {a+b}\) | \(\ge\) | \(\displaystyle \dfrac 3 2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \frac {a+b+c} {b+c} + \frac {a+b+c} {a+c} + \frac {a+b+c} {a+b}\) | \(\ge\) | \(\displaystyle \frac 9 2\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by adding 3 | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \frac {a+b+c} {b+c} + \frac {a+b+c} {a+c} + \frac {a+b+c} {a+b}\) | \(\ge\) | \(\displaystyle \frac {9 \left(a+b+c\right)} {\left(b+c\right) + \left(a+c\right) + \left(a+b\right)}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | $\dfrac {a+b+c} {\left(b+c\right) + \left(a+c\right) + \left(a+b\right)} = \dfrac 1 2$ | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \frac {\frac {1} {b+c} + \frac {1} {a+c} + \frac {1} {a+b} } {3}\) | \(\ge\) | \(\displaystyle \frac {3} {\left(b+c\right) + \left(a+c\right) + \left(a+b\right)}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | by dividing by $3\left(a+b+c\right)$ |
These are the arithmetic mean and the harmonic mean of $\dfrac {1} {b+c}$, $\dfrac {1} {a+c}$ and $\dfrac {1} {a+b}$.
The arithmetic mean is never less than the harmonic mean, so the last inequality is true, thus Nesbitt's inequality holds.
$\blacksquare$
Find out whether this originated with Cecil J. Nesbitt or another Nesbitt.
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