Prove $\sin^n 2x+(\sin^n x - \cos^n x)^2\le 1$ in POTW #393

[anemone]

Here is this week's POTW:

-----

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

-----

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!

 

[anemone]

No one answered last week's POTW.

Here is a suggested solution from other:

Spoiler

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

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

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

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 $s^n c^n$.