# inequality with numbers--when its true?

Help me please to understand when the inequality true.

Let $n<N,$ where $n, N$ are natural numbers.

For which $n$ and $N$ the following is true $$n^{2n+1}\leq N^{N+1}?$$

Thank you.

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You can get a numeric solution by taking logs: $(2n+1) \ln n=(N+1) \ln N$. Given $N$, this is a one-dimensional root finding problem that yields to any reasonable technique. – Ross Millikan Dec 5 '12 at 22:43
...and given monotonicity of $(2n+1)\ln n$, the root $n_*$ has the useful property that $n<n_* \Rightarrow n^{2n+1} <N^N$ and $n>n_*\Rightarrow n^{2n+1}>N^N$. Newton's method starting with $n_0=\sqrt{N}$ might give something useful. – user12477 Dec 5 '12 at 22:53

The inequality holds if $n\leq N^{1/2}$ and if $N^{1/2}\leq N-\frac12$:
Assuming these hypotheses, we get $n^2\leq N$ and so $$n^{2n}\leq N^{N^{1/2}}\leq N^{N-\frac12}.$$ Multiplying through once more by $n$, $$n^{2n+1}\leq nN^{N-\frac12}\leq N^{1/2}N^{N-\frac12}=N^N.$$
Since $N$ is a natural number, $N^{1/2}\leq N-\frac12$ if and only if $N\geq 4$. So we can say that the inequality is satisfied IF $$n\leq\sqrt{N} \quad \hbox{and}\quad N\geq4.$$ This is only a sufficient condition: I doubt that it is also necessary.
For $N=1,2,3$, we can check directly that only $n=1$ satisfies the inequality in those cases.
You're correct that $n<\sqrt{N}$ isn't necessary. $N=5, n=3$ works, but obviously $3>\sqrt{5}$. – Rick Decker Dec 5 '12 at 22:23
And for $N=25, n \approx 14.965$ – Ross Millikan Dec 5 '12 at 22:40