Say n (>1) is the smallest integer such that for some minimal point ,
,
then
On the other hand the critical points for the function are
They are the points in such that
(note that by our finding above)
But
Thus we get a contradiction and we actually proved that
This is a bit stronger statement.
Thanks Sudharaka for the direction. I thought this problem needs more advanced tech to solve.
Dear elim,
We have to show that for any positive integer n and
When n=1,
Therefore the expression is true for n=1
Suppose the expression is true for n=p where
Then,
Now,
Therefore,
Hence,
Therefore our the expression is true for
Hence by mathamatical induction the expression is true for all
Hope this will help you.