Nesbitt's Inequality/Proof 1

Theorem

Let $a$, $b$ and $c$ be (strictly) positive real numbers.

Then:

$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$

Proof

\(\ds \frac a {b + c} + \frac b {a + c} + \frac c {a + b}\)

\(\ge\)

\(\ds \dfrac 3 2\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}\)

\(\ge\)

\(\ds \frac 9 2\)

by adding $3$

\(\ds \leadstoandfrom \ \ \)

\(\ds \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}\)

\(\ge\)

\(\ds \frac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} }\)

as $\dfrac {a + b + c} {\paren {b + c} + \paren {a + c} + \paren {a + b} } = \dfrac 1 2$

\(\ds \leadstoandfrom \ \ \)

\(\ds \frac {\frac 1 {b + c} + \frac 1 {a + c} + \frac 1 {a + b} } 3\)

\(\ge\)

\(\ds \frac 3 {\paren {b + c} + \paren {a + c} + \paren {a + b} }\)

dividing by $3 \paren {a + b + c}$

These are the arithmetic mean and the harmonic mean of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$.

From AM-HM Inequality the last inequality is true.

Thus Nesbitt's Inequality holds.

$\blacksquare$

Source of Name

This entry was named for Alfred Mortimer Nesbitt.