Which term is bigger? $\sqrt[102]{101}$ or $\sqrt[100]{100}$ Which term is bigger? $\sqrt[102]{101}$ or $\sqrt[100]{100}$
I tried AM-GM but didn't succeed. 
 A: If we raise both sides of $\sqrt[102]{101}$ and $\sqrt[100]{100}$ to the $100$ power we are then comparing $101^{100/102}$ and $100$. Or in other words $101^{50/51}$ and $100$. Raise both to the $51$ power and we're comparing $101^{50}$ and $100^{51}$.
Thus it's equivalent to compare $101^{50}$ and $100^{51}$.  
Thus it's equivalent to compare $\left(\frac{101}{100}\right)^{50}$ and $100$.
Now  $\left(\frac{101}{100}\right)^{50}=(1+\frac1{100})^{50}$
Using the binomial theorem this is
$$\sum_{n=0}^{50}{50\choose n}\left(\frac{1}{100}\right)^{50-n}$$
Which is
$$\sum_{n=0}^{50}\frac{50!}{n!(50-n)!}\left(\frac{1}{100}\right)^{50-n}$$
$$=\sum_{n=0}^{50}\frac{50\cdot49\cdots(n+1)}{(50-n)!}\left(\frac{1}{100}\right)^{50-n}$$
It is clear that $\frac{50\cdot49\cdots(n+1)}{100^{50-n}}<1$ since the top is $50-n$ numbers less than $100$ and the bottom is $50-n$ copies of $100$.
Thus
$$=\sum_{n=0}^{50}\frac{50\cdot49\cdots(n+1)}{(50-n)!}\left(\frac{1}{100}\right)^{50-n}$$
$$<50 < 100$$.
Thus $101^{50}$ is less than $100^{51}$.  Thus $\sqrt[102]{101}$ is less than $\sqrt[100]{100}$
A: $101^{50} = \sum_{k=0}^{50} \binom{50}{k} 100^{50-k}$.
${101^{50} \over 100^{51} } = {1 \over 100}  \sum_{k=0}^{50} \binom{50}{k} {1 \over 100^k} = {1 \over 100} (1+ {1 \over 100})^{50} = {1 \over 100} (1+ {{1 \over 2} \over 50})^{50} \le {1 \over 100} \sqrt{e} <1$.
A: let $100=n$ then we have to prove that $$(n+1)^{1/(n+2)}<n^{1/n}$$ this is equivalent to $$\left(1+\frac{1}{n}\right)^n<n^2$$ this is true for $n=100$ since the left-hand side has the limit $e$
A: $$\sqrt[102]{101}=(101)^{\frac{1}{102}}=\left(\left(101\right)^{\frac{25}{51}}\right)^{\frac{1}{50}}=\sqrt[50]{(101)^{\frac{25}{51}}}$$ 
$$\sqrt[100]{100}=\sqrt[50]{10}=(10)^{\frac{1}{50}}$$
So:
$$\sqrt[50]{(101)^{\frac{25}{51}}}<\sqrt[50]{10}$$
$$\sqrt[102]{101}<\sqrt[100]{100}$$
A: Let $R$ be the unknown relation. $R \in \{<,=,>\}$.
$$101^{1/102}\, R \,100^{1/100}$$
$$101^{100}\, R \,100^{102} $$
We did this by raising both sides to the $102 \times 100$.
$$(1+100)^{100} = 100^{100}(\frac{1}{100}+1)^{100} < 100^{100}e$$
When compared to $100^{102}$, the answer is clear.
A: $\sqrt[100]{100}>\sqrt[102]{101}\iff 100^{102}>101^{100}\iff 100^{51}>101^{50}$
$\iff 100>\left(\frac{101}{100}\right)^{50}$. This inequality gives for $x,r>0$ that $(1+x)^r\le e^{rx}$, so   
$\left(1+0.01\right)^{50}\le e^{50\cdot 0.01}=\sqrt{e}\ll 100$.   
Proof of the inequality is simple: $\left((1+x)^{\frac{1}{x}}\right)^{rx}\le e^{rx}$.
A: Consider function $f(x)=x^{1/x}$,
$f'(x)=x^{1/x-2}(1-\ln(x))$, 
so function $f$ is decreasing on the interval $(\mathrm{e},\infty)$,
thus 
$100^{1/100}>101^{1/101}>101^{1/102}$.
