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 $\displaystyle 0<r<1$ be a real number. Let $\displaystyle a,b>0$. Then $\displaystyle -1<r-1<0$ thus $\displaystyle (a+x)^{r-1} \leq x^{r-1}$ for $\displaystyle 0\leq x\leq b$. This means $\displaystyle \int_0^b (a+x)^{r-1} dx \leq \int_0^b x^{r-1} dx \implies (a+b)^r \leq a^r+b^r$.
Now if $\displaystyle a > b > 0 \implies (a-b)>0 \mbox{ and }b>0$. Using the above result it means $\displaystyle a^r = ((a-b)+b)^r < (a-b)^r + b^r \implies a^r - b^r < (a-b)^r$. Thus, if $\displaystyle n\geq 2$ then $\displaystyle 0< 1/n < 1$ and the above result holds.