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Let $ABC$ be an acute angle triangle.

Show that :

\begin{equation} \sum _{cyc}(\sin2B+\sin2C)^{2} \sin A \leq 12 \sin A \sin B \sin C \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(*)\end{equation}


My try:

I think one can start with applying the addition and subtraction formulas(maybe)! Then he/she has:

\begin{equation}(\sin2B+\sin2C)^2\sin A = 4\sin^2(B+C)\cos^2(B-C)\sin A\end{equation}

Since $A+B+C = 180$ then latest equation equals to:

\begin{equation} = 4\sin^3 A\cos^2(B-C)\end{equation}

But I have no ideas for continue :(


Edited($1/27/2014$) I used labbhattacharjee idea to writing the last equation in new form:

$4\sin^3A\cos(B-C) = 4\sin^2A\sin(B+C)\cos(B-C)$ $$=2\sin^2A(\sin2B + \sin2C)=(1-\cos2A)(\sin2B+\sin2C)$$ $$=(\sin2B+\sin2C)-\sin2B\cos2A-\sin2C\cos2A$$

Then we can write:

$$\sum_{cyc} 4 \sin^3A\cos(B-C)$$

$$=\sum_{cyc}(\sin 2B+\sin 2C) - \sum_{cyc}\sin2B\cos2A - \sum_{cyc}\sin2C\cos2A$$

$$=2\sum_{cyc} \sin2A-\sum_{cyc}\sin2B\cos2A-\sum_{cyc}\sin2A\cos2B$$

$$=2\sum_{cyc} \sin2A - \sum_{cyc}(\sin2B\cos2A+\sin2A\cos2B)$$

$$=2\sum_{cyc}\sin(2A)- \sum_{cyc}\sin(2A+2B)$$

$$=2 \sum_{cyc}\sin2A + \sum_{cyc} \sin2C$$

$$= 3(\sin2A +\sin2B + \sin2C)$$

and by using equality proved in this page we conclude that (look at equation $(*)$):

$$ \sum _{cyc}(\sin2B+\sin2C)^{2} \sin A = 12 \sin A \sin B \sin C \leq 12 \sin A \sin B \sin C $$

and the equality becomes true when $ABC$ is a equilateral.

Is there any other solution?

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  • $\begingroup$ You might want to show some perspective in the form of thoughts and attempts on this as an additional paragraph by way of edit, otherwise it might be closed for a lack of detail. $\endgroup$ – J. W. Perry Jan 25 '14 at 7:04
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    $\begingroup$ @J.W.Perry Ok. I'll edit the post in a few minutes to render my endeavor. $\endgroup$ – Zia Jan 25 '14 at 7:07
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    $\begingroup$ One observation : Using AM $\ge GM\sum 4\sin^3A\ge3\prod 4\sin A$ Now, using this(math.stackexchange.com/questions/154505/…), $\sum 4\sin^3A\ge 3(\sum \sin2A)$ $\endgroup$ – lab bhattacharjee Jan 27 '14 at 16:20
  • $\begingroup$ @labbhattacharjee See my additional paragraph. $\endgroup$ – Zia Jan 27 '14 at 20:14

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