Show that $((x+iy)(\frac 1 x +\frac i y ))^2$ is always less or equal to -4? I'm trying to figure out all the different (calculus) ways to solve this one.
Given the point $((x+iy)(\frac 1 x +\frac i  y ))^2$, 
where $x,y>0$ are positive real numbers,
a) show that it lies on the negative axis in the complex plane;
b) show that it is always less or equal to $-4$. (for $x,y>0$ and real) .
So for a), I simplified it, and it simply became $-\left(\frac {x^2 + y^2} {yx} \right)^2$. Since $x,y>0$ and real, the number must be negative and real.
For b) however I'm a bit stuck (maybe the mean value theorem?).  I'm interested in all the ways to solve this one so feel free to be creative.
 A: Expanding the term of interest reveals 
$$\left((x+iy)\left(\frac1x+i\frac1y\right)\right)^2=\left((x+iy)\left(\frac1x+i\frac1y\right)\right)^2=-\left(\frac xy+\frac yx\right)^2$$
which is real and negative for $x>0$ and $y>0$.

Next, note that the function $f(t)=t+t^{-1}\ge 2$ for $t>0$ (Take the derivative and set it to $0$ to see that $f$ is a minimum at $t=1$ where $f(1)=2$).  
Therefore, since $x/y= (y/x)^{-1}$, it is easy to see that for $x>0$ and $y>0$, we have $f(x/y)\le 2$ and hence
$$\frac xy+\frac yx\ge 2$$
whence we see that 
$$-\left(\frac xy+\frac yx\right)^2\le -4$$
as was to be shown!
A: you still need to show that
$-(\frac {x^2 + y^2}{xy})^2 < -4\\
|\frac {x^2 + y^2}{xy}| > 2\\
x^2 + y^2 > 2|xy|\\
x^2 - 2|xy| + y^2 > 0\\
(|x| - |y|)^2 > 0$
Which is indeed always true.
I missed the note that $x,y>0$  It makes things a little simpler as the absolute value above becomes necessary.
A: $$(x+iy)\left(\frac1x+\frac1yi\right)=\left(\frac xy+\frac yx\right)i\implies\left[(x+iy)\left(\frac1x+\frac1yi\right)\right]^2=-\left(\frac xy+\frac yx\right)^2=$$
$$=-\frac{x^2}{y^2}-2-\frac{y^2}{x^2}=-\frac{(x^2+y^2)^2}{x^2y^2}$$
and that's point (a). 
For (b):
$$-\frac{(x^2+y^2)^2}{x^2y^2}\le-4\iff(x^2+y^2)^2\ge4x^2y^2\iff(x^2-y^2)^2\ge0\;\;\color{green}{\checkmark}$$
A: Suppose
$$-(\frac {x^2 + y^2} {yx}  )^2>-4$$
then 
$$(\frac {x^2 + y^2} {yx}  )^2<4$$
$$(\frac {x^2 + y^2} {yx}  )^2<2$$
$$x^2 + y^2<2xy$$
$$x^2 + y^2-2xy<0$$
$$(x-y)^2<0$$
which is absurd.
