Proof the following inequality: $ (x -\frac{x^k-r}{kx^{k-1}})^k \geq r $ Given that $k \geq 2$, $r \geq 1$, $x > 0$ and $x^k \geq r$, I am supposed to proof that
$(x -\frac{x^k-r}{kx^{k-1}})^k \geq r $. It also says, Bernoulli's inequality might me helpful, but so far I had no success.
I only came this far:
$(x -\frac{x^k-r}{kx^{k-1}}) = \frac{kx^{k} - x^k +r}{kx^{k-1}} = \frac{kx^{k} - x^k +r}{kx^{k-1}} = \frac{(k-1)x^{k} +r}{kx^{k-1}}  $
And that if  $(\frac{(k-1)x^{k} +r}{kx^{k-1}})^k \geq (\frac{k-1+r}{k})^k = (1 + \frac{r-1}{k})^k $ was true then I could use Bernoulli, but I have no Idea how to actually show that this ist true. So some help would be appreciated.

Bernoulli's inequality does help:
$$
\left(x -\frac{x^k-r}{kx^{k-1}}\right)^k =
x^k \left(1 -\frac{x^k-r}{kx^{k}}\right)^k 
\ge x^k \left(1-k\frac{x^k-r}{kx^{k}}\right) = r \, .
$$
The inequality can be applied because
$$
 -\frac{x^k-r}{kx^{k}} \ge -\frac 1k > -1 \, .
$$
This works for all positive $x$, not only if $x^k \ge r$. It shows that the Newton iteration for finding solutions of $x^k - r = 0$ produces only values which are all greater than (or equal to) $\sqrt[k]{r}$.
Another way to obtain this result is to use that the function $f(x) =  x^k-1$ is convex:
$$
 x^* = x -\frac{x^k-r}{kx^{k-1}} = x - \frac{f(x)}{f'(x)}
$$
implies that
$$
 f(x^*) \ge f(x) + (x^*-x) f'(x) = 0 \, .
$$
