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 $\paren{\alpha_1, \alpha_2, \alpha_3}$-space consisting of the points inside the triangle with the vertices $P_1(\pi, 0,0), P_2(0, \pi, 0), P_3(0,0, \pi)$.

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 $0 \leq F<1$ 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

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

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}=F\left(\cot \alpha_1-\alpha_1^{-1}\right), $ and, as $F \neq 0$, (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 $f^{\prime}=-\sin ^{-2} \alpha+\alpha^{-2}<0$;

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

hence (7) implies

$\ds

\alpha_1=\alpha_2=\alpha_3(=\pi / 3) \text {. } $ 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 argumentation 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 $F \leq k$ 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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