What is the correct way to compare to algebraic quantities? If I have two algebraic quantities, what is the correct way to determine if they are equal, or if not which is greater than the other?
For example, if I have $\frac{2d}{\sqrt{s^2-c^2}}$ and $\frac{2ds}{s^2-c^2}$, and I want to prove $\frac{2d}{\sqrt{s^2-c^2}} < \frac{2ds}{s^2-c^2}$ for all positive $d$, $s$, and $c$ I know that I can set them "equal" and treat them as an equation like so:
$$\frac{2d}{\sqrt{s^2-c^2}}=\frac{2ds}{s^2-c^2}$$
$$\frac{2d}{\sqrt{s^2-c^2}}\cdot\left(s^2-c^2\right) = \frac{2ds}{s^2-c^2}\cdot\left(s^2-c^2\right)$$
$$2d\sqrt{s^2-c^2}=2ds$$
$$\sqrt{s^2-c^2}=s$$
$$s^2-c^2=s^2$$
$$0<c^2$$ because we are given $c>0$.
Then you can replace all the equals signs all the way up with less thans since I didn't multiply by $-1$. This doesn't strike me as particularly formal, so what is the formal way to accomplish this, and in addition, is this even a valid line of reasoning?
 A: $$\begin{align}\frac{2ds}{s^2-c^2}-\frac{2d}{\sqrt{s^2-c^2}}&=\frac{2d}{\sqrt{s^2-c^2}}\left(\frac{s}{\sqrt{s^2-c^2}}-1\right)\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s-\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}}\\&=\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-(s^2-c^2)}{\sqrt{s^2-c^2}(s+\sqrt{s^2-c^2})}\\&=\frac{2dc^2}{(s^2-c^2)(s+\sqrt{s^2-c^2})}\\&\gt 0.\end{align}$$
Or easily, $$\frac{2ds}{s^2-c^2}\gt\frac{2d}{\sqrt{s^2-c^2}}\iff \frac{2ds}{s^2-c^2}\cdot \frac{s^2-c^2}{2d}\gt\frac{2d}{\sqrt{s^2-c^2}}\cdot \frac{s^2-c^2}{2d}\iff s\gt\sqrt{s^2-c^2}$$
$$\iff s^2\gt s^2-c^2\iff c^2\gt 0.$$
Since $c^2\gt 0$ is true, the first inequality is also true.
A: After messing around by doing scratch work like what you've done, I like to prove what I originally wanted by using a single chain of operations:
\begin{align*}
\frac{2d}{\sqrt{s^2-c^2}}
&= \frac{2d}{\sqrt{s^2-c^2}} \cdot \frac{\sqrt{s^2-c^2}}{\sqrt{s^2-c^2}} \\
&= \frac{2d\sqrt{s^2-c^2}}{s^2-c^2} \\
&< \frac{2d\sqrt{s^2 + 0}}{s^2-c^2} &\text{since } c > 0 \iff c^2 > 0 \iff -c^2 < 0\\
&= \frac{2ds}{s^2-c^2} &\text{since } s > 0\\
\end{align*}
