$\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!
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.