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. 

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 $\sqrt[100]{100}$
