An Elementary Algebra (Ratios & Proportions) Problem

Question

Let $a,b,c,d$ be positive real numbers in Continued Proportion (i.e., $\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$), then show that

$$d-a \ge 3(c-b).$$ or $$d -a = 3 + (some\ algebraic\ expression) $$

Answer 1

Four positive real numbers in a geometric progression can be written in a form: $$ (a, aq, aq^2, aq^3) \text{,} $$ where $a, q \in \mathbb{R}_+$. Suppose there are some values $a\in \mathbb{R}_+,q \in \mathbb{R}_+ \setminus 1$ such that $$ \frac{|aq^3 - a|}{|aq^2 - aq|} < 3 \text{.} $$ It is equivalent to $$ \frac{|a(q^2+q+1)(q-1)|}{|aq(q-1)|} < 3 \iff q^2-2q+1 < 0 \iff (q-1)^2 < 0 \text{.} $$ What is a condradiction, from which we conclude that for all $a \in \mathbb{R}_+,q \in \mathbb{R}_+ \setminus 1$ $$ \frac{|aq^3 - a|}{|aq^2 - aq|} \geq 3 \text{.} $$ If $q=1$ then $$ 0-0\geq 3 \cdot 0 \text{.} $$

Answer 2

Note that $(a,b,c,d)=(a,ra,r^2a,r^3a)$.

Case $r\ge 1:$ $$r^2+r+1 \ge 3$$

Case $r< 1:$ $$r^2+r+1 < 3$$ In either case, $$(r^2+r+1)(r-1)a \ge 3(r-1)a$$ $$r^3a-a \ge 3(ra-a)$$ $$d-a \ge 3(c-b)$$