Is $\max\left(\frac{A}{B},\frac{C}{D}\right) \ge \frac{\max(A,C)}{\max(B,D)}$ 
Is $\max\left(\frac{A}{B},\frac{C}{D}\right) \ge \frac{\max(A,C)}{\max(B,D)}$?
Given that $A,B,C,D>0$.
What about $\frac{A}{B},\frac{C}{D}>1$.

Is there a better bound for the left hand side of the equation?
 A: The inequality is true when $A,B,C,D$ are positive (no need to assume
$\frac{A}{B}>1$ or things like that). Let $\mu=\frac{{\sf max}(A,C)}{{\sf max}(B,D)}$
and $M={\sf max}\bigg(\frac{A}{B},\frac{C}{D}\bigg)$.
If $A\leq C$, then
$$
\mu=\frac{C}{{\sf max}(B,D)} \leq \frac{C}{D} \leq M.
$$
If $C\leq A$, then
$$
\mu=\frac{A}{{\sf max}(B,D)} \leq \frac{A}{B} \leq M.
$$
A: What if $A = D = 1$ and $B=C=-1$? Then, $LHS=-1$ and $RHS=1$.
A: $\frac{5}{6}$ and $\frac{3}{2}$  This one is wrong.  Thanks to Shakespeare for pointing it out.

The inequality holds: Let's assume (without loss of generality) that $\frac{a}{b}\ge \frac{c}{d}$.  
Case 1: $a\ge c$ and $b\ge d$. Then in the inequality of the problem, we need $\frac{a}{b}\ge \frac{a}{b}$. True.
Case 2: $a\ge c$ and $b<d$ Then in the inequality of the problem we need $\frac{a}{b}\ge\frac{a}{d}$.  This is true in this case since $b<d$.
Case 3: $a<c$ and $b\ge d$.  But then $ad<bc$, contradicting $\frac{a}{b}\ge\frac{c}{d}$
Case 4: $a<c$ and $b<d$.  Then in the inequality of the problem we need $\frac{a}{b}\ge\frac{c}{d}$.  But that's true because of our WLOG assumption.
A: Hint
$\frac{A}{\max(B,D)} \leq \frac AB \leq \max (\frac AB, \frac CD)$
$\frac{C}{\max(B,D)} \leq \frac CD \leq \max (\frac AB, \frac CD)$
