MHB Can You Solve This Trigonometric Inequality?

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The discussion presents a trigonometric inequality to prove: $\sin^n 2x + (\sin^n x - \cos^n x)^2 \le 1$. Participants are encouraged to engage with the Problem of the Week (POTW) and refer to the guidelines for submissions. There is a note that last week's POTW went unanswered, highlighting a lack of participation. A suggested solution from another user is mentioned, indicating that some attempts to solve the problem are being shared. The thread emphasizes the importance of community involvement in solving mathematical challenges.
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Here is this week's POTW:

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Prove that $\sin^n 2x+(\sin^n x - \cos^n x)^2\le 1$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered last week's POTW.

Here is a suggested solution from other:
Let $s=\sin x$ and $c=\cos x$, then the left-hand side expression becomes

$2^ns^nc^n+(s^n-c^n)^2=(2^n-2)s^nc^n+s^{2n}+c^{2n}$

while the right-hand side expression becomes

$$1=(s^2+c^2)^n=s^{2n}+c^{2n}+\sum_{i=1}^{n-1}{n \choose i}s^{2n-2i}c^{2i}$$

Now, we have to show that $$(2^n-2)s^nc^n\le \sum_{i=1}^{n-1}{n \choose i}s^{2n-2i}c^{2i}$$.

But that is immediate from AM-GM inequality that applies to the $2^n-2$ terms of $s^n c^n$.

Note that there are the same number of terms $s^{2n-2i}c^{2i}$ and $s^{2i}c^{2n-2i}$ and the product of each pair is $s^{2n}c^{2n}$. Hence the geometric mean is $s^n c^n$.
 
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