Apostol Theorem 1.35 (Every nonnegative real number has a unique nonnegative square root) 
In the above theorem I am not able to understand inequalities in the two marked boxes. 
To be specific in the first inequality $0<c<b$, I don't understand why we have $c<b$ when $c=\frac{1}{2}(b+\frac{a}{b})$. In the second inequality how do we get $(b+c)^2<b^2+3bc$?
 A: Look just before the first box. You have
$$
c = b - \frac{b^2 - a}{2b},
$$
so that $c$ is "$b$ minus a little something", as long as that little something is actually positive!
Now $b > 0$, so the denominator is positive, so all we need to show, to prove that $c < b$, is that the numerator is positive. But the first words of the paragraph are "suppose $b^2 > a$," and that makes $b^2 - a > 0$. 
For the second, we know that 
$$
(b+c)^2 = b^2 + 2bc + c^2
$$
so if we can show that $c^2 < bc$, we can rewrite this as
$$
(b+c)^2 = b^2 + 2bc + c^2 < b^2 + 2bc + bc = b^2 + 3bc.
$$
So...why is $c^2 < bc$? Well, $c$ is positive, so that's the same as asking "Why is $c < b$?", and the answer is in the previous sentence: we picked a number $c < b$ ...
I applaud you for double checking that you understand every single line of the proof; that'll help you craft good proofs in the future. 
A: At the beginning of that paragraph $c=b - \frac{(b^2-a)}{2b}$ which is clearly $<b$ as we substract from $b$ a positive number (as $b^2-a>0$ in this subcase) 
and that $b - \frac{(b^2-a)}{2b}$ equals $\frac{1}{2}(b + \frac{a}{b})$ is a simple fraction manipulation:
$$c= b - \frac{(b^2-a)}{2b}= \frac{2b^2}{2b} - \frac{(b^2-a)}{2b} = \frac{b^2+a}{2b} = \frac{1}{2}(\frac{b^2+a}{b})= \frac12(b + \frac{a}{b}) >0 $$
For the second we choose $0<c<b$ so 
$$b^2 + c(2b+ c) < b^2 + c(2b+b) = b^2 + 3bc$$
whiel $c < \frac{a-b^2}{3b}$ implies $3bc < a-b^2$ and so the rest of the line is explained too.
A: *

*$c<b\iff  \frac12\Bigl(b+\cfrac ab\Bigr)<b\iff \frac 12 \dfrac ab<\frac 12b\iff a<b^2$, which is the hypothesis in this case.

*$(b+c)^2=b^2+c(2b+c)<b^2+3bc\iff c(2b+c)<3bc\iff2b+c<3b\iff c<b$, the hypothesis in this case again.

