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Problem Of The Week #393 Nov 20th, 2019

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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

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

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 $s^n c^n$.
 
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