Nesbitt's Inequality

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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 1

\(\ds \frac a {b + c} + \frac b {a + c} + \frac c {a + b}\) \(\ge\) \(\ds \dfrac 3 2\)
\(\ds \leadstoandfrom \ \ \) \(\ds \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}\) \(\ge\) \(\ds \frac 9 2\) 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.

$\blacksquare$


Proof 2

Let:

\(\ds S\) \(=\) \(\ds \frac a {b + c} + \frac b {c + a} + \frac c {a + b}\)
\(\ds M\) \(=\) \(\ds \frac b {b + c} + \frac c {c + a} + \frac a {a + b}\)
\(\ds N\) \(=\) \(\ds \frac c {b + c} + \frac a {c + a} + \frac b {a + b}\)


Then:

\(\text {(1)}: \quad\) \(\ds M + N\) \(=\) \(\ds \frac {b + c} {b + c} + \frac {c + a} {c + a} + \frac {a + b} {a + b}\)
\(\ds \) \(=\) \(\ds 3\)
\(\text {(2)}: \quad\) \(\ds S + M\) \(=\) \(\ds \frac {a + b} {b + c} + \frac {b + c} {c + a} + \frac {c + a} {a + b}\)
\(\ds \) \(\ge\) \(\ds 3 \sqrt [3] {\frac {a + b} {b + c} \cdot \frac {b + c} {c + a} \cdot \frac {c + a} {a + b} }\) Cauchy's Mean Theorem
\(\ds \) \(=\) \(\ds 3\)
\(\text {(3)}: \quad\) \(\ds N + S\) \(=\) \(\ds \frac {c + a} {b + c} + \frac {a + b} {c + a} + \frac {b + c} {a + b}\)
\(\ds \) \(\ge\) \(\ds 3 \sqrt [3] {\frac {c + a} {b + c} \cdot \frac {a + b} {c + a} \cdot \frac {b + c} {a + b} }\) Cauchy's Mean Theorem
\(\ds \) \(=\) \(\ds 3\)

From $(2)$ and $(3)$:

$(4): \quad M + N + 2 S \ge 6$

From $(1)$ and $(4)$:

\(\ds 2 S\) \(\ge\) \(\ds 3\)
\(\ds \leadsto \ \ \) \(\ds S\) \(\ge\) \(\ds \frac 3 2\)

$\blacksquare$


Source of Name

This entry was named for Alfred Mortimer Nesbitt.


Sources