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Math Help - Try to prove this inequality

  1. #1
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    Try to prove this inequality

     \forall n>1, \sqrt{1+\sqrt{n+\sqrt{n^2+...\sqrt{n^n}}}}<n

    I have tried to solve it by recurrence with classical inequalities on square root but didn't get any result.
    Do you have any clue?

    Thanks!
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  2. #2
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    Opalg's Avatar
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    To get at something like this, you have to start with the innermost square root and work your way outwards.

    So start by looking at n^{n-1} + \sqrt{n^n} = n^{n-1} + n^{n/2}. Since n/2<n-1, it follows that n^{n-1} + \sqrt{n^n} < 2n^{n-1}.

    Now take the next square root: \sqrt{n^{n-1} + \sqrt{n^n}} < \sqrt2\,n^{(n-1)/2}. Since (n-1)/2<n-2, we can repeat the previous argument and get n^{n-2} + \sqrt{n^{n-1} + \sqrt{n^n}} < (1+\sqrt2)n^{n-2}.

    Take the next square root: \sqrt{n^{n-2} + \sqrt{n^{n-1} + \sqrt{n^n}}} < \sqrt{1+\sqrt2}\,n^{(n-2)/2} < \sqrt2\,n^{n-3}. (The last bit follows from the facts that \sqrt{1+\sqrt2}<2 and (n-2)/2<n-3.)

    Continuing in this way (using induction if you want to prove it properly), you see that for k=1,2,...,n-2, \sqrt{n^{n-k} + \sqrt{n^{n-k+1} +\ldots \sqrt{n^n}}} < \sqrt{1+\sqrt2}\,n^{(n-k)/2} < \sqrt2\,n^{n-k-1}.

    For k=n-2, this says that \sqrt{n^2 + \sqrt{n^3 +\ldots \sqrt{n^n}}} < \sqrt2\,n. Therefore \sqrt{n+\sqrt{n^2 + \sqrt{n^3 +\ldots \sqrt{n^n}}}} < 2\sqrt n . (You need to be a bit careful at this stage, because when k=n-1 it's no longer true that (n-k)/2<n-k-1.)

    Finally, \sqrt{1+\sqrt{n + \sqrt{n^2 +\ldots \sqrt{n^n}}}} < \sqrt{1+2\sqrt n}. It's then easy enough to check that the right-hand side is less than n.
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  3. #3
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    Thank you !
    This was not obvious to guess.
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