Abi-Khuzam Inequality

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Theorem

Let $\triangle ABC$ be a triangle.

Then:

$\sin A \cdot \sin B \cdot \sin C \le k A \cdot B \cdot C$

where:

$A, B, C$ are measured in radians
$k = \paren {\dfrac {3 \sqrt 3} {2 \pi} }^3 \approx 0 \cdotp 56559 \, 56245 \ldots$

Condition for Equality

$A=B=C$

Proof


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For $0 < \alpha_i < \pi, \alpha_1 + \alpha_2 + \alpha_3 = \pi$ we consider the function:

$\map F {\alpha_1, \alpha_2, \alpha_3} = \dfrac {\sin \alpha_1 \sin \alpha_2 \sin \alpha_3} {\alpha_1 \alpha_2 \alpha_3}$

$F$ is defined on the region $G$ in the $\tuple {\alpha_1, \alpha_2, \alpha_3}$-space consisting of the points inside the triangle with the vertices:

\(\ds P_1\) \(:=\) \(\ds \tuple {\pi, 0, 0}\)
\(\ds P_2\) \(:=\) \(\ds \tuple {0, \pi, 0}\)
\(\ds P_3\) \(:=\) \(\ds \tuple {0, 0, \pi}\)

We define $F$ on the boundary of the triangle by:

$\map F {0, \alpha_2, \alpha_3} = \dfrac {\sin \alpha_2 \sin \alpha_3} {\alpha_2 \alpha_3}, \alpha_2 \neq 0, \alpha_3 \neq 0, \alpha_2 + \alpha_3 = \pi$

and analogously:

$\map F {\alpha_1, 0, \alpha_3}$ and $\map F{\alpha_1, \alpha_2, 0}$

furthermore:

$\map F {\pi, 0, 0} = \map F {0, \pi, 0} = \map F {0, 0, \pi} = 0$

$F$ is now defined on a closed region $\bar G$; it is continuous and differentiable on $\bar G$.

Moreover, as $0 \le F < 1$ there is (at least) one point in $\bar G$ where $F$ has its maximum value.

By the usual procedure, in view of $\alpha_1 + \alpha_2 + \alpha_3 = \pi$, a maximum satisfies:

\(\text {(6)}: \quad\) \(\ds \frac {\partial F} {\partial \alpha_1}\) \(=\) \(\ds \frac {\partial F} {\partial \alpha_2}\) \(\ds = \frac {\partial F} {\partial \alpha_3} \paren {= \lambda}\)

In $G$ we have:

$\ds \frac {\partial F} {\partial \alpha_1} = \frac {\sin \alpha_2 \sin \alpha_3} {\alpha_2 \alpha_3} \cdot \frac {\alpha_1 \cos \alpha_1 - \sin \alpha_1} {\alpha_1^2} = \map F {\cot \alpha_1 - \alpha_1^{-1} }$

and, as $F \ne 0$, $(6)$ implies:

\(\text {(7)}: \quad\) \(\ds \cot \alpha_1 - \alpha_1^{-1}\) \(=\) \(\ds \cot \alpha_2 - \alpha_2^{-1}\) \(\ds = \cot \alpha_3 - \alpha_3^{-1}\)

For $f = \cot \alpha - \alpha^{-1}$ we obtain:

$f' = -\sin^{-2} \alpha + \alpha^{-2} < 0$

$f$ is therefore a decreasing function of $\alpha$ (we have $0 > f > -\infty$)

hence $(7)$ implies

$\ds \alpha_1 = \alpha_2 = \alpha_3 (= \pi / 3)$

in this point we have $F = k$.

We must verify whether larger values appear on the boundary of $\bar G$.

Between $P_2$ and $P_3$ yields

$\ds F = \frac {\sin \alpha_2 \sin \alpha_3} {\alpha_2 \alpha_3}, \alpha_2 + \alpha_3 = \pi$

and by an argument analogous to the former, but now with two factors instead of three, it follows that for the maximum on $P_2 P_3$ we have $\alpha_2 = \alpha_3 = \pi / 2$ and $F = 4 / \pi^2$, that is less than $k$.

Hence $F \le k$ on $\bar G$, which concludes the proof.

$\blacksquare$


Source of Name

This entry was named for Faruk Fuad Abi-Khuzam.


Sources

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