Which quantity is greater, $\frac{x^2}{y+\frac1y}$ or $\frac{y^2}{x+\frac1x}$? $x \gt y$, $ xy \neq 0$
A= $ x^2\over {y+{1\over y}}$
B= $ y^2\over {x+{1\over x}}$
Options:
1) Quantity A is greater.
2) Quantity B is greater.
3) The two quantities are equal.
4) The relationship cannot be determined from the given information.
By taking $x=2, y=1 $, I get $A\gt B$ , Thus options 2 and 3 are eliminated.
By taking$ x=2, y=-1$ , I get $B\gt A$ , Thus option 1 is eliminated.
So answer is option 4.
But I am not satisfied with this solution by taking particular values of $x$ and $y$.
Is there any other method to deal with this question?
What should be proper tag for this?
I think it should be comparision but that is not available in tag list.So please edit it.
 A: Suppose I compute the difference $A-B$ and see if it is always of one sign.
$$\begin{align}A - B &=\frac{x^2}{y+1/y} - \frac{y^2}{x+1/x}\\
&=\frac{yx^2}{y^2+1} - \frac{xy^2}{x^2+1}\\
&=\frac{xy\left(x(x^2+1)-y(y^2+1)\right)}{(x^2+1)(y^2+1)}\\
&=\frac{xy(x-y)(x^2+xy+y^2+1)}{(x^2+1)(y^2+1)}\\
&=\frac{xy(x-y)\left((x+y/2)^2+3y^2/4+1\right)}{(x^2+1)(y^2+1)}\end{align}$$
Every term in the last expression is positive but the term $xy.$ Therefore $$A > B \text{ if and only if } xy > 0.$$
A: If D really is the true answer, then supplying counterexamples to every other claim is the only way to prove it. And there is no shame in doing it that way.
A: A nice approach is as follows: note first of all that $x + 1/x$ is positive if $x$ is positive and negative if $x$ is negative. Now, if $x$ and $y$ have the same sign, then
$$
A > B \iff\\
\frac{x^2}{y+1/y} > \frac{y^2}{x+1/x} \iff \\
x^2(x + 1/x) > y^2(y + 1/y) \iff \\
x^3 + x > y^3 + y
$$
Note, however, that $f(x) = x^3 + x$ is an increasing function.  So, $x^3 + x > y^3 + y \iff x > y$.  So, $A > B \iff x > y$
On the other hand, if $x$ and $y$ have opposite signs, then
$$
A > B \iff\\
\frac{x^2}{y+1/y} > \frac{y^2}{x+1/x} \iff \\
x^2(x + 1/x) < y^2(y + 1/y) \iff \\
x^3 + x < y^3 + y \iff\\
x < y
$$
So, it suffices to take any positive values $x<y$, then the values $-x,y$.
A: Informally, I think you should be able see that as $x$ and $y$ both increase towards positive infinity, A is larger. As $x$ and $y$ both approach negative infinity, A is smaller.
You can show this by letting $(x, y)>0$, and $x = ay$ for $a > 1$. Then the opposite for the negative side 
