Prove or Disprove - simple inequality conjecture for ratios of real numbers Please exhibit a proof or counterexample for the following claim.
Let $x,y,a,b \in \mathbb{R}$ and $0 < c < 1$. Then  $$|x/y - 1| < c \text{  and  } |a/b - 1| < c \iff |ay/bx - 1| < c.$$ 
If the bi-conditional does not hold, does either direction hold?
What have I tried? Practically nothing. I am very sick yet still working and would like to verify this for a program I am writing. Sorry, no energy for fiddling with absolute value inequalities :/ You can browse my other questions and see that I usually give attempts and approaches, but right now I just need to move on with this.

counterexample for $\implies$: If $c=1-10^{-10}$, $a/b = 10^{-1}$, and $x/y = 10^{-5}$ then $\frac{ay}{bx} = 10^4$ which is very far from $1$.
counterexample for $\impliedby$: If $a/b = \frac{1}{4}$ and $x/y = \frac{1}{2}$ then $\frac{ay}{bx} = \frac{1}{2}$ which is within $2/3$ of $1$, but $|a/b-1| > 2/3$.
