<s>[INST] Proof of $n^{1/n} - 1 \le \sqrt{\frac 2n}$ by induction using binomial formula Using $$(a+b)^{n} =  \sum_{i=0}^n {n \choose k} a^{n-k} b^{K}$$
prove that 
$$n^{1/n} - 1 \le \sqrt{\frac 2n}$$ for n= 2,3,4.... 
I know the first step is to set $$ n^{\frac 1n} = 1 + x $$ for some x>0 and then raise both sides to the n, but I'm lost after that.
 [/INST] If $x=n^{1/n}-1$, then $n=(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\cdots\ge 1+\frac{n(n-1)}{2} x^2$ for $n\ge 2$.
Then $n-1\ge \frac{n(n-1)}{2} x^2\implies 1\ge \frac{n}{2}x^2\implies x^2\le\frac{2}{n}\implies x\le\sqrt{\frac{2}{n}}$
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