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Help me please to understand when the inequality true.

Let $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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    ...and given monotonicity of $(2n+1)\ln n$, the root $n_*$ has the useful property that n and n>n_*\Rightarrow n^{2n+1}>N^N. Newton's method starting with $n_0=\sqrt{N}$ might give something useful.2012-12-05

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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.

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    And for $N=25, n \approx 14.965$2012-12-05