A fraction problem $$a = x + \frac1x \\b =y + \frac1y \\ c = xy +\frac1{xy} $$ 
Express $c$ in terms of $a$ and $b$
 A: Note that $c$ is not uniquely specified by $a$ and $b$, since solving $a=x+\frac{1}{x}$ for $x$ yields two solutions which are reciprocals of each other, and applying $x\mapsto\frac{1}{x}$ or $y\mapsto\frac{1}{y}$ to $xy+\frac{1}{xy}$ yields $\frac{x}{y}+\frac{y}{x}$. However, applying either substitution again gives back $xy+\frac{1}{xy}$, so $c$ can take on either of those values.
Let $c_1=xy+\frac{1}{xy}$ and $c_2=\frac{x}{y}+\frac{y}{x}$. One can check that
$$c_1+c_2=ab$$
and
$$c_1c_2=a^2+b^2-4.$$
Use the quadratic formula to solve for the possible values of $c$.
A: Multiply all things by the LCD to remove fractions.
$$ax=x^2+1$$
$$by=y^2+1$$
$$cxy=x^2y^2+1$$
Multiply the first two equations to get
$$abxy=x^2y^2+1+x^2+y^2=c+x^2+y^2$$
Adding the first two equations, we get
$$ax+by=x^2+y^2+2$$
Combining these two lines,
$$abxy=c+ax+by-2$$
At this point, I'd just go back and solve for $x,y$ using the quadratic formula,
$$c+a\left(\frac{a\pm\sqrt{a^2-4}}2\right)+b\left(\frac{b\pm\sqrt{b^2-4}}2\right)-2=ab\left(\frac{a\pm\sqrt{a^2-4}}2\right)\left(\frac{b\pm\sqrt{b^2-4}}2\right)$$
Not the most beautiful thing in the world, but at least you don't have to square the thing or anything messy like that.
A: $a=x+1/x\iff (x^2-a x+1=0\land x\ne 0)\iff 0\ne x=(a\pm \sqrt {a^2-1})/2.$ Similarly $0\ne y=(b\pm \sqrt {b^2-1})/2.$ Now compute all  possible values of  $x y +1/x y. $ There are at most 2 values.
