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Question: Prove that $n^n>\left(\dfrac{n+1}{2}\right)^{n+1}$ for all positive integer $n>1$.

I could not understand what should be the initial approach.

Primo Raj
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    Applying weighted AM $\ge$ GM, we have \begin{align} & \left(\dfrac{1 \times \dfrac{1}{1} + n \times \frac{1}{n}}{1+n} \right)^{1+n}> \dfrac{1}{1^1 n^n}\ \implies & \left(\dfrac{2}{1+n} \right)^{1+n}> \dfrac{1}{1^1 n^n}\ \implies & n^n>\left(\dfrac{n+1}{2}\right)^{n+1} \end{align} – MKS Nov 29 '19 at 02:42

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We want to show the $2$s at the bottom are enough to defeat the $+1$s at the top. The effect of the $2$s is easy. Just write

$$ \left(\dfrac{n+1}{2}\right)^{n+1} = \frac{1}{2^{n+1}}(n+1)^{n+1}$$ We'd like a bound like $(n+1)^{n+1} < Cn^n$ for some $C < 2^{n+1}$.

$$(n+1)^{n+1}= (n+1)(n+1)^{n}$$

To bound $(n+1)^n$ use the binomial expansion, round each $n^i$ up to $n^n$, then use how the coefficients add up to $2^n$. This should complete the proof.

Daron
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