$\displaystyle \dfrac{5}{-4} < \dfrac{5}{4}$
If you were to cross-multiply, but keep the inequality as less than, you would get
$\displaystyle 20 = 5(4) < 5(-4) = -20$, which is clearly false.
Now, if you have an expression like
$\displaystyle \dfrac{f(x)}{g(x)} < \dfrac{h(x)}{j(x)}$, you must be certain of whether it is positive or negative. So, you can say the following:
$\displaystyle f(x)j(x) < g(x)h(x)$ if $\displaystyle g(x)j(x)>0$
$\displaystyle f(x)j(x) > g(x)h(x)$ if $\displaystyle g(x)j(x)<0$
Example 1:
$\displaystyle \dfrac{5}{(x+1)^2} < \dfrac{7x}{5}$
Since $\displaystyle 5(x+1)^2>0$ for all $\displaystyle x$, you know $\displaystyle 5(5) < 7x(x+1)^2$
Example 2:
$\displaystyle \dfrac{5}{-4} < \dfrac{7x^3}{1-(x+2)^2}$
Since $\displaystyle -4[1-(x+2)^2] = 4[(x+2)^2-1] \ge 4[(0+2)^2-1] = 4(4-1) = 12>0$, you know that $\displaystyle 5[1-(x+2)^2] < -7x^3(-4)$
Example 3:
$\displaystyle \dfrac{5}{-4} < \dfrac{7x^3}{(x+1)^2}$
Since $\displaystyle -4(x+1)^2 < 0$ for all $\displaystyle x$, you know $\displaystyle 5(x+1)^2 > 7x^3(-4)$ (note the inequality changed direction because of the single negative).