<s>[INST] Ellipse: Most direct algebraic demonstration that $\left|x\right|$ is maximal when $y=0$ Yes, I know this is the equation of an ellipse. 
This is the starting point of the problem:
$$
r_1+r_2=2a,
$$
where
$$r_1\equiv\sqrt{(x+c)^2+y^2}; r_2\equiv\sqrt{(x-c)^2+y^2}.$$
I want to establish certain properties prior to deriving a parametric expression for the ellipse, or even the "standard form",
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
$$
I can do all of this with a couple tacks, a string and a pencil, but
I want to show things algebraically. Preferably in a way that is directly applicable to all second-degree plane curves.
The first property I wish to establish is the range of $x$ values.
I know from experience that the answer is going to be that the extreme
values of $x$ will occur when $y=0$. The question that I am asking
in this post is: what is the most direct (algebraic) way to show that $\left|x\right|$
is greatest when $y=0$? 
I had hoped to establish that fact, and then use another (absolute value)
method of determining the actual values of $x$ when $y=0$. But, the best
I cam up with is this very long-winded demonstration.
First some definitions: Let $a,c,x,y\in\mathbb{R}$ where $a>c>0$
are constants, and $x$ and $y$ are variables such that, for $r_1\equiv\sqrt{(x+c)^2+y^2}$
and $r_2\equiv\sqrt{(x-c)^2+y^2}$ we have the equation
$$
r_1+r_2=2a.
$$
The degenerate case has been omitted.
I will presume the greatest value to be $\left|x\right|=a$, and take
the case of $x=a$. We square our equation
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(a+c)^2+y^2+2r_1 r_2+(a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2.
$$
So
$$
r_1(2a-r_1)=a^2-c^2-y^2
$$
$$
=2a\sqrt{(a+c)^2+y^2}-((a+c)^2+y^2)=a^2-c^2-y^2.
$$
Transpose, cancel and simplify
$$
2a\sqrt{(a+c)^2+y^2}=2a^2+2ac,
$$
$$
r_1=\sqrt{(a+c)^2+y^2}=a+c.
$$
And, therefore
$$
r_2=2a-r_1=a-c.
$$
Now, by definition we have
$$
\sqrt{\left(a+c\right)^2}\equiv\left|a+c\right|=r_1
$$
and
$$
\sqrt{(a-c)^2}\equiv\left|a-c\right|=r_2.
$$
If $y\ne0$ while $x=a$, then 
$$
r_1=\sqrt{(a+c)^2+y^2}=\left|a+c\right|+\delta_1,
$$
and 
$$
r_{2}=\sqrt{(a-c)^2+y^2}=\left|a-c\right|+\delta_2,
$$
where both $\delta_1>0$ and $\delta_2>0.$
Since $a>c>0,$ we have $r_1+r_2=2a+\delta_1+\delta_2$, in
contradiction to the equation we seek to satisfy. So $x=a\implies y=0.$ 
If $x>a$, let $\delta_3=x-a>0.$ 
So 
$$
r_1+r_2=a+c+\delta_3+a-c+\delta_3>2a;
$$
another contradiction.
Thus far we have dertermined that $x=a\implies y=0,$ and for any
$y$ that $x\le a$.
If we set $x=-a,$ then
$$
(r_1+r_2)^2=4a^2
$$
$$
=r_1^2+2r_1 r_2+r_2^2
$$
$$
=(-a+c)^2+y^2+2r_1 r_2+(-a-c)^2+y^2
$$
$$
=2(a^2+c^2+y^2)+2r_1 r_2
$$
$$
r_{1}\left(2a-r_{1}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}-\left(\left(-a+c\right)^{2}+y^{2}\right)=a^{2}-c^{2}-y^{2}
$$
$$
2a\sqrt{\left(-a+c\right)^{2}+y^{2}}=2a^{2}-2ac
$$
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c>0.
$$
$$
r_{2}=2a-r_{1}=a+c.
$$
Since we now have 
$$
r_{1}=\sqrt{\left(a-c\right)^{2}+y^{2}}=a-c,
$$
and
$$
r_{2}=\sqrt{\left(a+c\right)^{2}+y^{2}}=a+c,
$$
the argument showing that $x=-a\implies y=0,$ is essentially the
same. Likewise for the argument that $x\ge-a.$
And that's just the $x$ range.
NB: I don't "need" any of this to fully characterize the ellipse.  My goal is to find multiple ways of examining the same construct.
 [/INST] There is a very simple and short answer.  Using the definition of monotonicity given in  Fundamentals of Mathematics, Volume 1: Foundations of Mathematics: The Real Number System and Algebra Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle, which is a monotonic function is such that
$$x<y\implies f\left(x\right)<f\left(y\right)$$
or such that
$$x<y\implies f\left(y\right)<f\left(x\right),$$
and using the results of the elementary theory of numbers developed in Chapter 1B-1 for exponents, the proof is straight forward.  That was the part I was missing.  The result is "intuitively obvious", but I did not have a theorem to apply.  In particular, I had no proof of the the blatantly obvious fact
$$\sqrt{(x-c)^2+y^2}\ge\sqrt{(x-c)^2}.$$
</s>