Cfrgtkky

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:
$$ 0leq cos^2{theta}leq 1 ;; text{ therefore } ;; 0leq frac{(x+y)^2}{4xy}leq 1 $$
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?

We have $sec^2 thetage 1$ for all $theta$ at which $sectheta$ 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 $(x+y)^2ge 4xy$, or equivalently that $x^2-2xy+y^2ge 0$. 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.

$$sec^2theta=frac{4xy}{(x+y)^2}impliestan^2theta=frac{4xy}{(x+y)^2}-1=-frac{(x+y)^2-4xy}{(x+y)^2}=-left(frac{x-y}{x+y}right)^2$$

$$ifftan^2theta+left(frac{x-y}{x+y}right)^2=0 (1)$$

For real $displaystyle theta,tan^2thetage0$

and for real $displaystyle x,y; left(frac{x-y}{x+y}right)^2ge0$

So, each has to be individually zero to satisfy $(1)$

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$$
mbox{Let's}quad r equiv {x over x + y}quadmbox{such that}quad
{4xy over pars{x + y}^{2}} = 4rpars{1 - r} =1 - pars{2r - 1}^{2} leq 1
$$

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

$4xy$ can be rearranged as $(x+y)^2-(x-y)^2$, and therefore $sec^2theta=frac{(x+y)^2-(x-y)^2}{(x+y)^2}=1-left ( frac{x-y}{x+y} right )^2$, 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 $-1le costhetale1Rightarrow cos^2thetale1Rightarrow sec^2thetage1$.

We have therefore established that $sec^2thetale1$ and also that $sec^2thetage1$; both of these conditions can be satisfied only if $sec^2theta=1$.

$Rightarrow 1-left ( frac{x-y}{x+y} right )^2=1Rightarrow left ( frac{x-y}{x+y} right )^2=0$

Now if $x$ and $y$ are not equal to $0$, then the denominator $x+yne 0 Rightarrow x=y$ and neither of which is equal to $0 blacksquare$

from jay
we can show that

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

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

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

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

$$x-y le 0$$

$$x=y$$