Suppose $a$, $b$, $c$, and $d$ are real numbers, $0<a<b$, and $d>0$. Prove that if $ac\geq bd$ then $c>d$.

Question

Not a duplicate of

Prove that if $a$, $b$, $c$, and $d$ are real numbers and $0 < a < b$ and $d > 0$ and $ac ≥ bd$ then $c > d$

This is exercise $3.1.11$ from the book How to Prove it by Velleman $(2^{nd}$ edition$)$:

Suppose $a$, $b$, $c$, and $d$ are real numbers, $0<a<b$, and $d>0$. Prove that if $ac\geq bd$ then $c>d$.

The solution to this problem is available in appendix $1$, but I was wondering about another possible solution:

Proof. We will prove the contrapositive. Suppose $c\leq d$. Since $0<a<b$, then multiplying $a$ by $c$ and $b$ by $d$ will preserve the direction of the inequality and so $ac<bd$ as required. Therefore if $ac\geq bd$ then $c>d$. $Q.E.D.$

Is the above proof valid?

Answer 1

I'm not happy with "then multiplying a by c and b by d will preserve the direction of the inequality". I'm not sure that's clear and I'm not sure you can claim that without justification.

You could prove that as a lemma.

Lemma: If $m < n$ and $0 < u \le v$ the $mu < nv$.

Pf: $mu < nu$ and $nu \le nv$ (as well as $mu \le mv$ and $mv < nv$ ) so by transitivity $mu < nv$.

Answer 2

If $c \le d \implies ac \le ad < bd \implies ac < bd$, contradiction. Thus $c > d$.