Try to prove this inequality

• Oct 26th 2007, 06:45 AM
josephnyu
Try to prove this inequality
$\forall n>1, \sqrt{1+\sqrt{n+\sqrt{n^2+...\sqrt{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!
• Oct 26th 2007, 09:38 AM
Opalg
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, 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, 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.)

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.)

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.
• Oct 26th 2007, 09:50 AM
josephnyu
Thank you !
This was not obvious to guess.