Assume for the sake of contradiction that this is not the case.
Let where .
Similarly define , where .
Let be the unique positive integer satisfying
What we wish to show is that:
Since both the RHS and LHS are positive, this is equivalent to:
So we just need to show that:
It suffices to show:
Thus, we want to prove that:
Since the RHS of the above equation is greater than , the result follows iff:
As a result, it suffices to show that:
The above inequality is true iff:
But, by repeating the above procedure the exponent of the "explodes" since is a fixed yet arbitrary positive integer it must be passed eventually.
This is a contradiction and our lemma is true.
5. Let be positive real numbers. Prove that
It's supposed to work so that we prove 1-4 and then 5. Unfortunately I can't show 2 and 3.
Suppose that the real numbers satisfy:
Determine the maximum of:
and when equality occurs
I'm not sure "Olympiad level" is something that can be said for inequalities--AM-GM and almost all the others you listed are extremely standard. Perhaps you mean the application is somehow nonstandard or more difficult to see.
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