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#### melese

##### Member

- Feb 24, 2012

- 27

inequality holds: $\displaystyle P\leq\frac{\sqrt3}{4}(abc)^{2/3}$

Find all triangles for which equality holds.

- Thread starter melese
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- Thread starter
- #1

- Feb 24, 2012

- 27

inequality holds: $\displaystyle P\leq\frac{\sqrt3}{4}(abc)^{2/3}$

Find all triangles for which equality holds.

- Thread starter
- #2

- Feb 24, 2012

- 27

solution:

To get an inequality in the right direction, we try to determine the maximum value $m$ of $\sin(\alpha)\sin(\beta)\sin(\gamma)$. Since $\sin$ increases in $[0,\pi/2]$, we have $\sin(\alpha)\leq\sin(\beta)\leq\sin(\gamma)$. So it's not difficult to see that we need to maximize $\alpha$, and that happens for $\alpha=\pi/3$ ($\alpha+\alpha+\alpha\leq\alpha+\beta+\gamma=\pi$). Now, obviously $\alpha=\beta=\gamma=\pi/3$, and hence $\displaystyle m=\sin(\pi/3)\sin(\pi/3)\sin(\pi/3)=(\frac{\sqrt3}{2})^3$.

Finally, $\displaystyle P\leq(\frac{1}{8}(abc)^2m)^{1/3}=\frac{\sqrt3}{4}(abc)^{2/3}$.

Equality occurs precisely for equilateral triangles.

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