True or False?

If choosing $x=1;y=-2$ , the inequalities $A, B, C$ are not true.

If choosing $x=4; y=-1$ , the inequality $E$ is not true.

We have: $x>y\Leftrightarrow \dfrac{x}{x^2y^2}>\dfrac{y}{x^2y^2} \Leftrightarrow \dfrac{1}{xy^2}>\dfrac{1}{x^2y}$

So, the inequality $\boxed{D}$ is always true.

divide the inequality by (xy)^2 we get option D

Due to a case of trivial inequality, for all real numbers n, ${ n }^{ 2 }\ge 0$

and we know that $x>y$

Thus ${ y }^{ 2 }<{ x }^{ 2 }$

And the reciprocal of the inequality ${ y }^{ 2 }<{ x }^{ 2 }$ will be $\frac { 1 }{ { y }^{ 2 } } >\frac { 1 }{ { x }^{ 2 } }$

And thus option D is correct

Here, Keep x=2 and y=3 Then check the possible solutions. Now,keep x=-2 and y=-3 and recheck whether your solution holds true here too.

Choosing x as 1 and y as -1 eliminates all bar D instantly.

If you multiply both side of inequality (D) by xy, then it will be reduced to inequality (A). Then (A) and (D) are the same, can I make this way?

Just choose x and y as 2 positive integers for easy calculations, then 2 negative integers such that x > y

For negative values you will see true nature of equations and find out that only option D gives correct output in both cases.