Question

If x and y are both negative show that x/y + y/x > 2

Answer 1

\[\large\rm \frac xy+\frac yx>2\]Multiply both sides by x,
(Since x is negative, the inequality flips),
\[\large\rm \frac{x^2}y + y<2x\]Multiply both sides by y,
(y is negative, inequality flips again),
\[\large\rm x^2+y^2>2xy\]Subtract,\[\large\rm x^2-2xy+y^2>0\]These terms can factor down into a perfect square,
\[\large\rm (x-y)^2>0\]And since a squared term is always positive (always greater than zero),
this holds true.

`If x and y are both negative show that x/y + y/x > 2`
To properly "show" that, we would want to apply all of these steps in reverse.

We would start with (x-y)^2>0.
Mention that it holds true because it's a square.
And show that it expands to the equation we're looking for.

Answer 2

THanks a lot . I always have a hard time with this type of question.

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @celticcat
If x and y are both negative show that x/y + y/x > 2
\(\color{#0cbb34}{\text{End of Quote}}\)


A Contradiction:

let \(x = y = \alpha, \qquad \alpha \ne 0\)

\( \implies \dfrac xy+\dfrac yx = 2 \implies \dfrac xy+\dfrac yx \ngtr 2\)

So, Poof(!), it's wrong - this is math. And there are an infinite number of ways that the original proposition fails :(

There is also a singularity at the Origin

IMHO: using algebra here, is fine but hardly inspired. And using kids' rules, about flipping the sign of the inequality is, ..... mmmm, interesting.

not that interested in doing so, but i reckon you could work the proof in one original direction without the a priori back-tracking that the **5-star** answer suggests 😂

occurred to me that you could write it as \(u + \dfrac{1}{u} \gt 2\), then plot it as a single variable problem: that is an interesting approach, maybe