Query regarding Theorem 1.21 in Baby Rudin In the proof of Theorem - 1.21 (pg-10) in Rudin's Principles of Mathematical Analysis
(Statement - For every real $x>0$ & every integer $n>0$, there is one & only one positive real $y$ s.t. $y^n = x$); 


*

*The author says - "Assume $y^n<x$. Choose h so that $0<h<1$ & $h<\frac{x-y^n}{n(y+1)^{n-1}}$" $\space\space\space\space\space\space\space\space\space\space\space$ How does he get the last inequality?

*In the second part, how does the author get this value of k when he says - "Assume $y^n>x$. Put $k=\frac{y^n-x}{ny^{n-1}}$" ?
Thanks in advance...
 A: if $y \lt x^{\frac1{n}}$ is an approximate solution to:
$$
f(t)=x-t^n=0
$$
then Newton's method would suggest
$$
y + \frac{x-y^n}{ny^{n-1}}
$$
as a better one. however 
$$
\left(y + \frac{x-y^n}{ny^{n-1}}\right)^n =y^n+ny^{n-1}( \frac{x-y^n}{ny^{n-1}})+\cdots \gt x
$$
which is of no use to an argument requiring a mototone increasing approach to a limit. the motive for replacing $y$ by $y+1$ in the denominator of the Newton approximation is to obtain an improved estimate that does not overshoot. proving this is the only remaining difficulty...
in the other case this problem does not arise, and the treatment Rudin gives is just the Newton method
A: Here is my interpretation of how Rudin argued the first point in your question (to contradict the case that $y^{n} < x$).
The idea is to find a $y^*$ that would lie in the "gap" of numbers created if $y$ were indeed less than the "true $\sqrt[n]{x}$". (Here $y$ refers to $y=\sup E$ where $E=\{t:t^n < x, t\in\mathbb{R^+}\}$ as defined by Rudin.)
In other words, we are trying to construct such a $y^*$ that has the two properties $(y^*)^n < x$ and $y^* > y$. The first property says that $y^* \in E$, the set which $y$ bounds above. The second property contradicts the fact that $y$ is an upper bound of $E$, since $y^* \not\le y$ and $y^* \in E$.
To construct $y^*$ we need to find a suitable $h$ such that $y^*=y+h$.
Intuitively, one would like to let $h$ be a positive quantity less than the difference $\sqrt[n]{x} - y$. However, we have not yet shown the existence of $\sqrt[n]{x}$ so this complicates our approach.
One alternative is to look at the difference $x - y^n$, which is a valid expression at this point in the proof. Graphically, this approach can be thought of as picking the position of $y+h$ on the horizontal axis based on the function value  $(y+h)^n$ on the vertical axis.
In other words, instead of specifying $h$ directly we will try to specify $h$ in terms of what $y+h$ maps to under exponentiation by $n$.
Hence, we look for an $h$ such that $(y+h)^n - y^n < x - y^n$.
(Again, think of comparing function values of the curve $f(t) = t^n$ on a graph. The left-hand side $(y+h)^n - y^n$ is some positive quantity smaller than the height of the vertical "gap" assumed to exist between $y^n$ and $x$ on the graph.)
(Since $f(t) = t^n$ is a strictly increasing function (for positive $t$), inequalities are preserved, so the hypothetical horizontal "gap" (in which we were originally interested) will correspond with a vertical "gap" after this transformation. Intuitively, (pretending $\sqrt[n]{x}$ is defined) we could write,
$$
\sqrt[n]{x} - y > 0 \iff \sqrt[n]{x} > y \iff f(\sqrt[n]{x}) > f(y) \iff \sqrt[n]{x}^n > y^n \iff x > y^n \iff x - y^n > 0
$$
with the following justifications for each of the double arrows (in order from left to right): (1) addition/subtraction by $y$, (2) $f$ is strictly increasing, (3) definition of $f$, (4) definition of $\sqrt[n]{x}$, (5) addition/subtraction by $y^n$.)
Using the observation Rudin made$^1$ we can inject another expression into the inequality to get:
$(y+h)^n - y^n < hn(y+h)^n < x - y^n$
(This "injection" is okay for now because we have technically not defined $h$ yet, but instead are still working backwards to specify $h$.)
The trouble with this inequality is that the middle expression contains $h$ inside a binomial term raised to the power of $n$, which makes it hard to algebraically isolate. I believe Rudin makes the convenient assumption at this point that $h$ is small — in other words $0 < h < 1$ — in order to complete his definition of $h$.
We can now inject another expression into the inequality to yield:
$(y+h)^n - y^n < hn(y+h)^n < hn(y+1)^n < x - y^n$
Isolating just the rightmost two expressions and rearranging gives the final inequality:
$h < \frac{x - y^n}{n(y+1)^n}$
End Note

*

*Rudin's observation:

$b^n - a^n = (b-a)(b^{n-1} + ab^{n-2} + ... + a^{n-2}b + a^{n-1})$
so
$b^n - a^n < (b-a)n(b^{n-1})$
in the case $0 < a < b$.
(Rudin let $b = y+h$ and $a=y$ in the proof.)
