Inequality problem, How to prove I have this question:

Assume:
$$y\in(3,4)\\ x\in(1,2)$$
Prove:
$$\frac16<\frac{y-x}{xy}<\frac34$$

How do I solve it? 
I tried to put the X's and Y's inside this equation:
$$\frac{y-x}{xy}$$
like so:
Open
I tried to check whether the inequality is true in those bounders.
My lecturer told me that it is not true and I need to make manipulations on 1<x<2 and 3<y<4 so that the outcome will be the inequality above.
I don't know how to solve it and appreciate any help.
Thanks.
(Please excuse my English - this is not my main language).
 A: Note that$$\frac{y-x}{xy}=\frac1x-\frac1y.$$Now, since $\frac1x<1$ and $\frac1y>\frac14$,$$\frac1x-\frac1y<1-\frac14=\frac34.$$And, since $\frac1x>\frac12$ and $\frac1y<\frac13$,$$\frac1x-\frac1y>\frac12-\frac13=\frac16.$$
A: Hint: $$\frac{y-x}{xy}=\frac1x-\frac1y$$
And $\frac1x\in\left(\frac12,1\right)$ and $\frac1y\in\left(\frac14,\frac13\right).$
A: A more mechanical method that can be used when you do not see any "slick" way to leverage the structure of the objective function...
First, we look for critical points on $(1,2)\times(3,4)$.  However, $\frac{\partial}{\partial x} \frac{y-x}{xy} = \frac{-1}{x^2} \neq 0$, $\frac{\partial}{\partial y} \frac{y-x}{xy} = \frac{1}{y^2} \neq 0$, so there are no local extrema on $(1,2)\times(3,4)$.  Consequently, all extrema are on the boundary of that region.  Notice that the objective function is continuous in $x$ and $y$ (away from points with $x = 0$ or $y = 0$, at which the objective function is undefined), so we may extend it to the boundary and search for extrema on the boundary to be unattained extrema for the objective function.
(Notice we don't have to calculate the following derivatives -- we already did so in the second line of the previous paragraph.  So on all four boundary components, we know the objective is strictly monotonic, giving no extrema on the interiors of the boundary components.)
When $x = 1$, $(\mathrm{d}/\mathrm{d}y) \ (y-1)/y = 1/y^2$, so the objective is strictly monotonically increasing on $[3,4]$.  Candidates for extrema are
$$  \frac{3-1}{3} = \frac{2}{3} \quad \text{and} \quad \frac{4-1}{4} = \frac{3}{4}  \text{.}  $$
When $x = 2$, $(\mathrm{d}/\mathrm{d}y) \ (y-2)/(2y) = 1/y^2$, so the objective is strictly monotonically increasing on $[3,4]$.  Candidates for extrema are
$$  \frac{3-2}{6} = \frac{1}{6} \quad \text{and} \quad \frac{4-2}{8} = \frac{1}{4}  \text{.}  $$
(The next two checks are useful in the general case, but are redundant here.  We already know there are no extrema on the interiors of the next two boundary components, and we've already calculated the values on the ends of those components, as we will see.)
When $y = 3$, $(\mathrm{d}/\mathrm{d}x) \ (3-x)/(3x) = -1/x^2$, so the objective is strictly monotonically decreasing on $[1,2]$.  Candidates for extrema are
$$  \frac{3-1}{3} = \frac{2}{3} \quad \text{and} \quad \frac{3-2}{6} = \frac{1}{6}  \text{.}  $$
When $y = 4$, $(\mathrm{d}/\mathrm{d}x) \ (4-x)/(4x) = -1/x^2$, so the objective is strictly monotonically decreasing on $[1,2]$.  Candidates for extrema are
$$  \frac{4-1}{4} = \frac{3}{4} \quad \text{and} \quad \frac{4-2}{8} = \frac{1}{4}  \text{.}  $$
So $1/6$ is an unattained lower bound and $3/4$ is an unattained upper bound of the objective.  That is,
$$  \frac{1}{6} < \frac{y-x}{xy} < \frac{3}{4}  \text{.}  $$
