Nesbitt's Inequality/Proof 2
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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
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.