Prove that $\sec^2{\theta}=(4xy)/(x+y)^2$ only when $x=y$ Show that the equation below is only possible when $x=y$
  $$ \sec^2{\theta}=\frac{4xy}{(x+y)^2}$$




The only way I can think of doing this is by rewriting it as 

$$ \cos^2{\theta}=\frac{(x+y)^2}{4xy} $$
then using some inequalities to prove it by using:
$$ 0\leq \cos^2{\theta}\leq 1 \;\; \text{ therefore } \;\; 0\leq \frac{(x+y)^2}{4xy}\leq 1 $$
But I have an aversion to using case-based solutions (checking for $x&gt;0$, $y&gt;0$ etc.) since I feel there must be a neater solution to  these kind of problems. So my question is: Is it possible to solve this and these sort of questions using techniques that don't involve checking numerous cases?
We have $\sec^2 \theta\ge 1$ for all $\theta$ at which $\sec\theta$ is defined.  So it is enough to show that $\frac{4xy}{(x+y)^2}\le 1$, with equality only when $x=y$. 

To show that $\frac{4xy}{(x+y)^2}\le 1$, we show equivalently that $(x+y)^2\ge 4xy$, or equivalently that $x^2-2xy+y^2\ge 0$. But this is clear, since $x^2-2xy+y^2=(x-y)^2$. And we have equality precisely when $x=y$. 

Remark: This is not very different from how you proposed to do things. There are no cases involved. And aversion to cases can be problematic. A consideration of cases (though not in this case) is often a natural approach.
from jay 
we can show that 

$$(x+y)^2 \le 4xy$$

$$x^2+y^2+2xy \le 4xy$$

$$x^2+y^2-2xy \le 0$$

$$(x-y)^2 \le 0$$

$$x-y \le 0$$

$$x=y$$