- #1
jobsism
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Here's a question:-
Prove that for all positive integers n,
[(n+1)/2]^n >= n!
And here's a funny proof for it:-
Assume to the contrary that, for all positive integers n,
[(n+1)/2]^n < n!
However, for n=2,
(3/2)^2 > 2!
Therefore, our assumption must be false.
And hence, for all positive integers n, [(n+1)/2]^n >= n!
Now, I know that this proof can't be correct, because I've seen the real proof, and it's a marvel, making use of algebraic inequalities, and the proof above simply seems too simple compared to it. But I wonder, what's the mistake with the above proof?
Prove that for all positive integers n,
[(n+1)/2]^n >= n!
And here's a funny proof for it:-
Assume to the contrary that, for all positive integers n,
[(n+1)/2]^n < n!
However, for n=2,
(3/2)^2 > 2!
Therefore, our assumption must be false.
And hence, for all positive integers n, [(n+1)/2]^n >= n!
Now, I know that this proof can't be correct, because I've seen the real proof, and it's a marvel, making use of algebraic inequalities, and the proof above simply seems too simple compared to it. But I wonder, what's the mistake with the above proof?