Proof of $ y^n=x$ Rudin From the image below the inequalities I’m trying to wrap my head around are labelled A and B.I understand here that Rudin was attempting to expose a contradiction in the argument that y^n<x,but what I don't understand is his procedure of derivation for A and B and why he chose A and B instead of another set of inequalities.

 A: B) isn't constructed from A)  B) is constructed from $y^n < x$ with A) in mind for a later goal.
Perhaps it would have been clearer he had done B) first.
$y^n <x$
So $\frac {x-y^n}{n(y+1)^{n-1} }> 0$.
Let $0 < h < \min(1,\frac {x-y^n}{n(y+1)^{n-1}})$
There's no question that we can make such an $h$. 
The only question is why on earth we are doing it.
Then .... we take a tea break and observe, seemingly apropos nothing, that for $0 < a < b$ it so happens that 
$(b^n -a^n) = (b-a)(b^{n-1} + b^{n-2}a + ...... +ba^{n-2} + a^{n-1})$
$< (b-a)(b^{n-1} + b^{n-2}b + .... + b*b^{n-2} + b^{n-1}) = (b-a)nb^{n-1}$.
Okay.... that is a true statement. We can't deny that.
But what that has to do with anything is unclear at this point.  
Then he puts those two together:
We let $a = y$ and $b = y + h$ where $h$ was that weird number we made for no apparent reason, then 
$b^n - a^n < (b-a)nb^{n-1}$ so
$(y+h)^n - y^n < h*n(y + h)^{n-1} $
$< hn(y+1)^{n-1}$ (because $h < 1)$
$< \frac {x-y^n}{n(y+1)^{n-1}}n(y+1)^{n-1}$ (because $h < \frac {x-y^n}{n(y+1)^{n-1}}$).
$=x -y^n$.
And at that point it becomes clear why we constructed that strange number and stated that strange fact.
Because they allow us to conclude:
$(y+h)^2 < x$ so $y+h \in E$  and $y + h > y$ so $y$ is not an upper bound of $E$.  
A: Slightly off topic, but related: Theorem 1.21 can alternatively be proven merely by demonstrating that $\;f(y) = y^n\;$ is both well defined and continuous at all $y>0.$ 
Remember, a function f is continous iff f "can be drawn without picking up the pencil".
A: Rudin does not provide motivation / insights for most proofs (like the one in question here). It is best to ditch such unmotivated proofs and instead start devising your own.
The crux of the proof here is to find an $h>0$ such that $(y+h) ^n<x$ given that $x, y$ are positive and $y^n<x$. This is not difficult. Just start with the assumption that $h<1$ (you can replace $1$ by any positive number like say $10001$, but $1$ simplifies the calculations).
And then consider our goal $$(y+h)^n<x$$ in bit detail. We have via binomial theorem $$(y+h) ^n=y^n+nhy^{n-1}+\dots+h^n<y^n+h(ny^{n-1} +\dots +1)$$ and our target inequality is achieved if the RHS of the above equation is less than $x$. Thus we need to have $$h(ny^{n-1}+\dots +1)<x-y^n$$ or $$h<\frac{x-y^n} {ny^{n-1}+\dots +1}$$ Since we have assumed $h<1$ we have finally $$0<h<\min\left(1,\frac{x-y^n}{ny^{n-1}+\dots +1}\right)$$
A: A. From the condition $0 \lt a \lt b$ it follows that $a^k \lt b^k$ and therefore:
$$b^{n-1}+b^{n-2}a+ \ldots + ba^{n-2}+a^{n-1} \lt b^{n-1}+b^{n-2}b+ \ldots + bb^{n-2}+b^{n-1}=n b^{n-1}$$
B. $\,h\,$ is simply chosen in such a way as to then use A in order to derive the contradiction that follows. Suppose you started with an arbitrary $\,h \in (0,1)\,$ and wrote the next inequalities up to $\,\lt hn(y+1)^{n-1}\,$, at which point you'd wonder what additional condition would be required for $\,h\,$ in order to derive the sought-after contradiction $\,\lt x - y^n\,$. Then B follows naturally.
