# Math Help - help with inequality

1. ## help with inequality

Can someone please show me how to prove this inequality:

Let a > b > 0 and let n be a natural number greater than or equal to 2. Then
a^(1/n) - b^(1/n) < (a-b)^(1/n).

Thank you for your help. It is greatly appreciated.

2. Originally Posted by jamesHADDY
Can someone please show me how to prove this inequality:

Let a > b > 0 and let n be a natural number greater than or equal to 2. Then
a^(1/n) - b^(1/n) < (a-b)^(1/n).

Thank you for your help. It is greatly appreciated.
We can prove something stronger. Let $0 be a real number. Let $a,b>0$. Then $-1 thus $(a+x)^{r-1} \leq x^{r-1}$ for $0\leq x\leq b$. This means $\int_0^b (a+x)^{r-1} dx \leq \int_0^b x^{r-1} dx \implies (a+b)^r \leq a^r+b^r$.

Now if $a > b > 0 \implies (a-b)>0 \mbox{ and }b>0$. Using the above result it means $a^r = ((a-b)+b)^r < (a-b)^r + b^r \implies a^r - b^r < (a-b)^r$. Thus, if $n\geq 2$ then $0< 1/n < 1$ and the above result holds.

3. Thank you for answering. Is there a little bit easier way to show this (possibly without integration)? Maybe with sequences or mean value theorem?