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

Question

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

Author of "How to Prove It" proved it by contrapositive. However, I've tried to use another approach:

Given that $d > 0$,

Let's rewrite $c$ as $c = dq$.

$0 < a < b \implies 0 < ad < bd$

for $adq > bd$ to hold true, $q$ must be larger than $1$, hence $c > d$.

Is it correct?

Answer 1

You only have that $adq\geq bd,$ not $>.$

It’s still true that $q>1,$ but in either case it is not clear exactly how you know that $q >1.$

But you could have extended your chain of inequalities like this:

$$ 0 < ad < bd \leq ac,$$

and from this you get $ad < ac.$ Then use the fact that $a>0.$

Answer 2

Since $ac \ge bd$, we can write: $$ac \ge bd \Longrightarrow 1 < \frac{b}{a} \le \frac{c}{d} \Longrightarrow 1 < \frac{c}{d} \Longrightarrow c > d$$

Answer 3

Let $c\leq d$.

Thus, $$ac-bd=a(c-d)+d(a-b)<0,$$ which is a contradiction.