For which $x, y\in\mathbb{R ^+}$ do we have $|xy-\frac{1}{xy}|\le|x-\frac{1}{x}|+|y-\frac{1}{y}|$? I need to find all $x, y\in\mathbb{R^+}$ such that the following inequality holds.
$$\Big| xy-\dfrac{1}{xy}\Big|\le\Big|x-\dfrac{1}{x}\Big|+\Big|y-\dfrac{1}{y}\Big|$$
If I substitute $x=2$ and $y=3$ clearly that inequality fails.
How can I attempt this question? Any hint?
Thank you.
 A: Since both members are positive, we can square and get this equivalent $$x^2y^2+\frac{1}{x^2y^2}-2\le x^2+\frac{1}{x^2}-2+y^2+\frac{1}{y^2}-2+2|x-\frac{1}{x}||y-\frac{1}{y}|$$
Then gettting all terms but the last one to the left, factorizing  and multipying by $x^2y^2$ we get 
$$ (x^2y^2+1)(x^2-1)(y^2-1) \le 2x^2y^2|x-\frac{1}{x}||y-\frac{1}{y}|  $$
From this we know that all $x,y$ satisfying $(x^2-1)(y^2-1) \le 0 $ are solution. For other values, we can square again and cancel repeted (positive) factors, we get $$(x^2y^2+1)^2 \le 4x^2y^2$$
And from this it is easy to deduce that 
$$(x^2y^2-1)^2 \le 0$$
what is to say, $(xy)^2=1$
I hope this is a little helpful!
New edit: actually, if $ x^2= \frac{1}{y^2}$ then it is also true that $(x^2-1)(y^2-1) \le 0 $. So $x,y$ solve the inequality if and only if they satisfy this last condition, which means that $|x| \le 1 $ and $|y| \ge 1 $ or viceversa
A: I found an answer for this problem. So I think it is better to publish it.
Without loss of generality assume that $x\ge y\ge 1.$
Now $$xy-\dfrac1{xy}\le \left(x-\dfrac1x\right)+\left(y-\dfrac1y\right)$$
$$xy-x-y+1\le\dfrac1{xy}-\dfrac1x-\dfrac1y+1$$
$$(x-1)(y-1)\le\left(1-\dfrac1x\right)\left(1-\dfrac1y\right).$$
This gives only solutions are $x=1$ or $y=1.$
