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Let $m,n\in\mathbb N$, $n>m$. Prove inequality $$n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$$

My work so far:

$$\sqrt[n]{n!}=\sqrt[n]{1\cdot2\cdot...\cdot n}\le\frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}2.$$ Then $n\sqrt[n]{n!}\le\frac{n(n+1)}{2}$

$$\sqrt[m]{m!}=\sqrt[m]{1\cdot2\cdot...\cdot m}\le\frac{1+2+...+m}{m}=\frac{m(m+1)}{2m}=\frac{m+1}2.$$ Then $m\sqrt[m]{m!}\le\frac{m(m+1)}{2}$

And $$\frac{n(n+1)}{2}-\frac{m(m+1)}{2}=\frac{(n−m)(n+m+1)}2.$$

But inequalities can be added, without subtract

Roman83
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  • If $n$ and $m$ are big enough, the LHS is close to $\frac{n^2-m^2}{e}$ by Stirling's approximation, and that is trivially less than $\frac{(n-m)(n+m)}{2}$. – Jack D'Aurizio Apr 20 '16 at 14:15

1 Answers1

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The result follows immediately by writing \begin{align} n\sqrt[n]{n!} &=n\sqrt[n]{m!\cdot(m+1)(m+2)\cdots n}\\ &=n\sqrt[n]{(\sqrt[m]{m!})^m\cdot(m+1)(m+2)\cdots n}\\ &\le m\sqrt[m]{m!}+(m+1)+(m+2)+\cdots+n\\ &=m\sqrt[m]{m!}+\frac{(n-m)(n+m+1)}{2}. \end{align}

Solumilkyu
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