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Decide which number is larger, $\sqrt[10000]{10001}$ or $\sqrt[9999]{10000}$. There's a hint: use Bernoulli inequality. So I've tried something like that $x=\sqrt[10000]{10001}$, $(\frac{1}{x})^{10000}=\frac{1}{10001}$ then $\frac{1}{10001}=((\frac{1}{x}-1)+1)^{10000}\geq 1+10000(\frac{1}{x}-1)$ which leads to $ x\geq \frac{10001}{10000}$. I can write down similar inequality with $y=\sqrt[9999]{10000}$ but It doesn't seem to help very much. Can you give me any hints?

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  • $\begingroup$ Hint: $(1+x)^{1/x}$ is decreasing. $\endgroup$ – Macavity Nov 15 '16 at 18:11
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Let $n=10000$, and let $$\lambda=\frac{n^{1/(n-1)}}{(n+1)^{1/n}},$$ then (as suggested by Daniel Fischer) we have $$\lambda^{n}=\frac{n^{\frac{n}{n-1}}}{n+1}=\frac{1}{n+1}(1+(n-1))^{n/(n-1)}>\frac{1}{n+1}(1+n)=1,$$ and thus $\lambda>1$.

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I just use the Newton Binomial Theorem, not the Bernoulli Inequality. Let $x=10000$. Then $$ \sqrt[10000]{10001}=(x+1)^{\frac{1}{x}},\sqrt[9999]{10000}=x^{\frac{1}{x-1}}. $$ which is equivalent to comparing $$ (x+1)^{x-1},x^{x} $$ or $$ \left(1+\frac{1}{x}\right)^{x},x+1. $$ By the Newton Binomial Theorem, one has $$ \left(1+\frac{1}{x}\right)^{x}=\sum_{k=0}^x\binom{x}{k}\frac{1}{x^k}=\sum_{k=0}^x\frac{x(x-1)\cdots(x-k+1)}{k!}\frac{1}{x^k}\le\sum_{k=0}^x\frac{1}{k!}<3< x+1. $$ So $$ (x+1)^{x-1}<x^{x}. $$

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Let $n=10000$, so that we are comparing $(n+1)^{1/n}$ with $n^{1/(n-1)}$. Let $r=n^{1/(n-1)}/(n+1)^{1/n}$ be the ratio. Then

$$r^{n(n-1)}={n^n\over(n+1)^{n-1}}={n+1\over\left(1+{1\over n}\right)^n}$$

If you know that $\left(1+{1\over n}\right)^n\approx e$ when $n$ is large, you're done: $r\gt1$, hence $10000^{1/9999}\gt10001^{1/10000}$. If you don't know about $e$, it's possible to prove by induction that $(1+{1\over n})^n\lt3$ for all $n\gt0$, which gives the same conclusion. (Remark: the standard approach for the induction is to prove the slightly stronger inequality $(1+{1\over n})^n\lt3-{1\over n}$. That little $-{1\over n}$ after the $3$ makes a world of difference.)

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