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 $x>0$, $y>0$ 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?
 A: 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. 
A: $$\sec^2\theta=\frac{4xy}{(x+y)^2}\implies\tan^2\theta=\frac{4xy}{(x+y)^2}-1=-\frac{(x+y)^2-4xy}{(x+y)^2}=-\left(\frac{x-y}{x+y}\right)^2$$
$$\iff\tan^2\theta+\left(\frac{x-y}{x+y}\right)^2=0\ \ \ \ (1)$$
For real $\displaystyle \theta,\tan^2\theta\ge0$
and for real $\displaystyle  x,y;  \left(\frac{x-y}{x+y}\right)^2\ge0$
So, each has to be individually zero to satisfy $(1)$
A: $\newcommand{\+}{^{\dagger}}
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$$
\mbox{Let's}\quad r \equiv {x \over x + y}\quad\mbox{such that}\quad
{4xy \over \pars{x + y}^{2}} = 4r\pars{1 - r} =1 - \pars{2r - 1}^{2} \leq 1
$$

Since $\ds{\sec^{2}\pars{\theta} \geq 1}$ the only solution occurs when $\ds{r = \half}$ which leads to $\color{#00f}{\large x = y}$.

A: $4xy$ can be rearranged as $(x+y)^2-(x-y)^2$, and therefore $\sec^2\theta=\frac{(x+y)^2-(x-y)^2}{(x+y)^2}=1-\left ( \frac{x-y}{x+y} \right )^2$, which is always less than or equal to $1$ because $\left ( \frac{x-y}{x+y} \right )^2$ is greater than or equal to $0$.
Recall that $-1\le \cos\theta\le1\Rightarrow \cos^2\theta\le1\Rightarrow \sec^2\theta\ge1$.
We have therefore established that $\sec^2\theta\le1$ and also that $\sec^2\theta\ge1$; both of these conditions can be satisfied only if $\sec^2\theta=1$.
$\Rightarrow 1-\left ( \frac{x-y}{x+y} \right )^2=1\Rightarrow \left ( \frac{x-y}{x+y} \right )^2=0$
Now if $x$ and $y$ are not equal to $0$, then the denominator $x+y\ne 0 \Rightarrow x=y$ and neither of which is equal to $0 \blacksquare$
A: from jay 
we can show that 
$$(x+y)^2 \le 4xy$$
$$x^2+y^2+2xy \le 4xy$$
$$x^2+y^2-2xy \le 0$$
$$(x-y)^2 \le 0$$
$$x-y \le 0$$
$$x=y$$
