Understanding Spivak's answer Chapter 1, 16(b) and the role of contradiction in a proof 
Using the fact that ${x}^{2} + 2xy + {y}^{2} = (x + y)^{2} \ge 0$ , show that $4{x}^{2} + 6xy + 4{y}^{2} > 0$ unless $x$ and $y$ are both $0$

This has been asked and answered here, but not in the spirit that Spivak has asked.
Spivak's answer:

The first equation implies that $$4{x}^{2} + 8xy + 4{y}^{2} \ge 0$$  Suppose we also had $$4{x}^{2} + 6xy + 4{y}^{2} \le 0$$ Subtracting the second from the first would give us $2xy \ge 0$. If neither $x$ nor $y$ is $0$, this means we must have $2xy>0$; But this implies that $4{x}^{2} + 6xy + 4{y}^{2} > 0$, a contradiction.
Moreover, it is clear that if one of $x$ and $y$ is $0$, but not the other then we also have $4{x}^{2} + 6xy + 4{y}^{2} > 0$

My analysis...
By subtracting the second from the first equation, we are able to isolate $xy$, and we show that $2xy \ge 0$. Now, to expand a little on Spivak, if
neither $x$ nor $y$ is $0$, then either $x,y > 0 $, or $x,y < 0$ (in order to obtain $2xy > 0$)
Applying that to $4{x}^{2} + 6xy + 4{y}^{2}$ gives us the desired answer that $4{x}^{2} + 6xy + 4{y}^{2} > 0$.
And, to finish off, if either $x$ or $y$ is $0$, we are left with $4y^{2} > 0$ or $4x^{2} > 0$
...and a question
Assuming the analysis is correct, I would like to understand more fully the issue of contradiction

Suppose we also had $4{x}^{2} + 6xy + 4{y}^{2} \le 0$


this means we must have $2xy>0$; But this implies that $4{x}^{2} + 6xy + 4{y}^{2} > 0$, a contradiction.

Does this contradiction imply the opposite is true? ie because ($4{x}^{2} + 6xy + 4{y}^{2} \le 0$) is false, then ($4{x}^{2} + 6xy + 4{y}^{2} > 0$) must be true? (Even though we go from $\le 0$ to $> 0$)
 A: 

Suppose we also had $4{x}^{2} + 6xy + 4{y}^{2} \le 0$


this means we must have $2xy>0$; But this implies that $4{x}^{2} + 6xy + 4{y}^{2} > 0$, a contradiction.

Does this contradiction imply the opposite is true? ie because
($4{x}^{2} + 6xy + 4{y}^{2} \le 0$) is false, then $(4{x}^{2} + 6xy + 4{y}^{2} > 0)$ must be true? (Even though we go from $\le 0$ to $> 0$)

Since our entire argument has been valid, the contradiction (absurdity or logical falsity) that has been derived must have arisen from the premise/assumption; in other words, our assumption must be false; in other words, its negation must be true.
I think calling the negation "opposite" is potentially ambiguous, and possibly why you ask the next question:

$(4{x}^{2} + 6xy + 4{y}^{2} > 0)$ must be true? (Even though we go from $\le 0$ to $> 0$)

The negation of $a>b$ is $a\leq b,$ while its "opposite" might be reasonably construed as $a<b.$
P.S. In the given exercise, the proof's premise/assumption is being contradicted, leading us to conclude that it is false. So, in fact, this proof would have been more succinctly cast as a contrapositive proof. The contradiction that is derived in a proof by contradiction does not in general involve the deliberately false assumption. More here.
A: OP may no longer be interested in this question, but perhaps this will be of use to someone someday.
There are some facts important to this proof, that are not explicitly stated.
The term $4x^2 + 6xy + 4y^2$ is by assumption a real number, and obeys the Trichotomy law. For any specific values of $x$ and $y$, the number $4x^2 + 6xy + 4y^2$ is either zero, positive, or negative.
The proof given by Spivak shows that this number cannot be negative, and can only be zero if both $x$ and $y$ are zero.
Since we know $4x^2 + 6xy + 4y^2$ obeys the Trichotomy law, we know it must be positive in all other cases.
(We can be even more explicit: $x$ and $y$ are real numbers by assumption.
By definition, the sum of a pair of real numbers is a real number. The product of a pair of real numbers is a real number. The expression $4x^2 + 6xy + 4y^2$ is composed of sums and products of this sort and so it is a real number. Real numbers follow the Trichotomy law...This sort of thing is unnecessary to include in most proofs, but it may have been helpful for Spivak to include it here in this early part of the book.)
