# Try to prove this inequality

• Oct 26th 2007, 05:45 AM
josephnyu
Try to prove this inequality
$\displaystyle \forall n>1, \sqrt{1+\sqrt{n+\sqrt{n^2+...\sqrt{n^n}}}}<n$

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!
• Oct 26th 2007, 08:38 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 $\displaystyle n^{n-1} + \sqrt{n^n} = n^{n-1} + n^{n/2}$. Since $\displaystyle n/2<n-1$, it follows that $\displaystyle n^{n-1} + \sqrt{n^n} < 2n^{n-1}$.

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

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

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

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

Finally, $\displaystyle \sqrt{1+\sqrt{n + \sqrt{n^2 +\ldots \sqrt{n^n}}}} < \sqrt{1+2\sqrt n}$. It's then easy enough to check that the right-hand side is less than n.
• Oct 26th 2007, 08:50 AM
josephnyu
Thank you !
This was not obvious to guess.