<s>[INST] 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 $(b+c)^2<b^2+3bc$?
 [/INST] 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.
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