<s>[INST] If $p, q, r$ and $s$ are in continued proportion. Then find the minimum value of $\frac{p-s}{q-r}.$ My approach: We have, $\frac pq=\frac qr=\frac rs.$ Which implies $q^2=pr$ and $r^2=qs.$ On dividing we get:
$\frac{q^2}{r^2}=\frac{pr}{qs}$
$\implies \frac{q^3}{r^3}=\frac{p}{s}$
$\implies \frac{q^3-r^3}{r^3}=\frac{p-s}{s}$
\begin{align}
\implies\frac{p-s}{q-r}&=\frac{s(q^2+qr+r^2)}{r^3}\\\\
&=\frac{s(q^2+qr+r^2)}{qs\cdot r},\text{ as }r^2=qs\\\\
&=\frac{q^2+qr+r^2}{qr}\quad ... (1) 
\end{align}
Now $(q-r) ^2\ge 0,$ true for all real $q$ and $r$. This gives
$q^2+r^2\ge 2qr\implies q^2+qr+r^2\ge 3qr.$
If $qr>0, $ then from $(1),$
$\min\left(\frac{p-s}{q-r}\right)=3.$
But I'm confused how to approach the same if $qr<0.$ Please suggest... (No additional informations were given in the question and I don't know the answer. I found this question on Quora, no one has answered it.) 
 [/INST] Let
$$\frac pq = \frac qr = \frac rs := a$$
We can express ratios in terms of $a$:
$$\frac pq = a \qquad \qquad \frac rq = \frac 1a \qquad \qquad \frac sq = \frac sr \frac rq = \frac{1}{a^2}$$
Hence, the expression becomes
$$\frac{p-s}{q-r} = \frac{\frac pq - \frac sq}{1 - \frac rq} = \frac{a - \frac{1}{a^2}}{1-\frac 1a} = \frac{a^3-1}{a^2-a} = \frac{a^2+a+1}{a}$$
assuming that $q \neq 0$, $a \neq 0$, and $a-1\neq 0$. Differentiating,
$$\frac{d}{da}\frac{a^2+a+1}{a}=0 \iff 1-\frac{1}{a^2} =0  \iff a = \pm 1$$
We compute
$$\frac{a^2+a+1}{a} \bigg|_{a=1} = 3 \qquad \qquad \frac{a^2+a+1}{a} \bigg|_{a=-1} = -1$$
Assuming that the ratio must be positive the minimum value is indeed $3$.
If the ratio is allowed to be negative, then clearly
$$\lim_{a\rightarrow -\infty} \frac{a^2+a+1}{a} = -\infty$$
so there is no minimum.
</s>