Question on the logic of a proof involving complex numbers The title is a bit vague, but really boils down to not the steps of a proof--I've been able to reason through what I know, via reviewing the solution and the suggestion in the text, that the steps are correct--but 'why' they'e correct.
I want to show that, for any complex number, $z$, $\sqrt{2} \lvert z \rvert \geq \text{Re $z$} + \text{Im $z$}$. The suggestion in the text is to manipulate this expression until we end up with $\left(\lvert x \rvert + \lvert y \rvert\right)^2 \geq 0$, which is necessarily true. 
I was able to find this, which concludes the proof. My question is, really, why this concludes the proof. I believe the idea is that, via manipulating the inequality, we conclude that this original inequality is true for all $x$ and $y$ that satisfy the inequality that the square of the sum of their absolute values is non-negative. Because that is necessarily true for any real numbers, this clearly holds for all reals and thus for all complex numbers, which are created from reals. 
The other idea I had was that we could retrace our steps to produce the original expression, but that would require reversing one step that required us to square both sides: we maintained the inequality sign because both sides of the inequality were positive, but it need not be true, unless I'm mistaken, that we end up with a 'positive' result via reversing the inequality, so we need not end up with our original result. 
I'd appreciate any insights on this problem. Thanks in advance. If necessary, I could write out the steps of the proof, especially if there is any merit to the second explanation from above. 
Revision: I have revised this question to include the steps of this proof. 
\begin{align*}
& \sqrt{2} \lvert z \rvert = \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert \text{Re $z$} \rvert + \lvert \text{Im $z$} \rvert \\
& \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert x \rvert + \lvert y \rvert \\
& 2(x^2 + y^2) \geq \left(\lvert x \rvert + \lvert y \rvert\right)^2 \\
& 2x^2 + 2y^2 \geq \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert + \lvert y \rvert^2 \\
& \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert y^2 \geq 0 \\
& \left(\lvert x \rvert - \lvert y \rvert\right)^2 \geq 0.
\end{align*}
 A: As long as the steps are reversable then "Conclusion $\iff .... \iff .... \iff$ something obviously true" is valid as we could just as easily do it the other way  "something obviously true $\iff ... \iff ... \iff $ Conclusion".
But the steps must be reversible, or at least "only if" statements.  They can not be "forward if" statments.
If you write the proof in the exact opposite order it is straigtforward and makes sense.
$\begin{align*}
& \left(\lvert x \rvert - \lvert y \rvert\right)^2 \geq 0\text{(which is obviously true)}\implies\\
& \lvert x \rvert^2 - 2 \lvert x \rvert \lvert y\rvert +|y^2| \geq 0 \implies\\
& 2x^2 + 2y^2 \geq \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert + \lvert y \rvert^2\implies \\
& 2(x^2 + y^2) \geq \left(\lvert x \rvert + \lvert y \rvert\right)^2 \implies\\
& \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert x \rvert + \lvert y \rvert \implies\\
& \sqrt{2} \lvert z \rvert = \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert \text{Re $z$} \rvert + \lvert \text{Im $z$} \rvert \\
\end{align*}$
It's just that every step feels like "where are you going with this?  How did you know to go in that direction?".  It's easier to follow the logic if we work backwards-- start with the conclusion and show how the conclusion must follow from an obviously true statement:
$\begin{align*}
& \sqrt{2} \lvert z \rvert = \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert \text{Re $z$} \rvert + \lvert \text{Im $z$} \rvert \Leftarrow\\
& \sqrt{2} \sqrt{x^2 + y^2} \geq \lvert x \rvert + \lvert y \rvert \Leftarrow\\
& 2(x^2 + y^2) \geq \left(\lvert x \rvert + \lvert y \rvert\right)^2 \Leftarrow\\
& 2x^2 + 2y^2 \geq \lvert x \rvert^2 + 2 \lvert x \rvert \lvert y \rvert + \lvert y \rvert^2 \Leftarrow\\
& \lvert x \rvert^2 - 2 \lvert x \rvert \lvert y \rvert +|y^2| \geq 0 \Leftarrow\\
& \left(\lvert x \rvert - \lvert y \rvert\right)^2 \geq 0\text{(which is obviously true)}.
\end{align*}$
