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Let $a, b \in [0,1]$, then for any $n$ $$a^{1/n}- b^{1/n}\leq (a-b)^{1/n} \leq a-b.$$

I did not seem to find a way of expanding this and see the result. I thought of Pascal's Triangle thing but did not help. Help please!

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Do you want to assume $a\geq b$? Otherwise, what does $(a-b)^{1/n}$ mean for $a<b$? – Zev Chonoles Dec 27 '12 at 6:21
I wonder if there is a nice way using precalculus (+1). The question is just nice. – user 1618033 Dec 27 '12 at 12:45
up vote 3 down vote accepted

Consider $f(t) = t^n - (t-1)^n$ for $t \geq 1$ and $n \geq 1$. We have $f'(t) = n \left( t^{n-1} - (t-1)^{n-1}\right) > 0$. Hence, $f(t)$ is an increasing function. Hence, we have $$f(t) \geq f(1) \implies t^{n} - (t-1)^n \geq 1 \,\,\,\,\,\, (\star)$$ i.e. $$(t-1)^n \leq t^n - 1$$ Now assuming $a\geq b$. Let $a = x^n$ and $b = y^n$. We now have $x>y$. This means $$x - y \leq (x^n - y^n)^{1/n} \implies (x-y)^n \leq x^n - y^n$$ Dividing by $y$, we get that $$(t-1)^n \leq t^n-1$$ which is true by $(\star)$.

However $(a-b)^{1/n} \leq (a-b)$ is incorrect since $(a-b) \in [0,1]$. And for $x \in [0,1]$, we have $x^n \leq x$ i.e. $x^{1/n} \geq x$. Hence, $(a-b)^{1/n} \geq (a-b)$

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