Prove that $\sec^2{\theta}=(4xy)/(x+y)^2$ only when $x=y$

Question

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

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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?

Answer 1

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.

Answer 2

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

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

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 $(1)$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \mbox{Let's}\quad r \equiv {x \over x + y}\quad\mbox{such that}\quad {4xy \over \pars{x + y}^{2}} = 4r\pars{1 - r} =1 - \pars{2r - 1}^{2} \leq 1 $$

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

Answer 4

from jay we can show that

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

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

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

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

$$x-y \le 0$$

$$x=y$$