Intuitive way of solving this inequality: $x^2>y^2 \wedge x>0 \Rightarrow x>y$

Question

I have to prove that $x^2>y^2 \wedge x>0 \Rightarrow x>y$.

I decided to do this: $x^2>y^2 \Leftrightarrow \sqrt{x^2}>\sqrt{y^2}\Leftrightarrow |x|>|y| \Leftrightarrow x>|y| \vee x<-|y|$, where the third equivalence is obtained by definition of absolute value. So, $x^2>y^2 \wedge x>0 \Rightarrow x>|y|$ And $|y|\geq y$, so $x^2>y^2 \wedge x>0 \Rightarrow x>y$.

My question is if is there a more intuitive way of solving this.

Any help would be appreciated.

Answer 1

$$x^2\gt y^2\implies 1\gt\left({y\over x}\right)^2\implies1\gt{y\over x}$$

with no condition on $x$. But if $x\gt0$, then

$$1\gt{y\over x}\implies x\gt y$$

Answer 2

Well, if $y \ge 0$, then since $x \mapsto x^2$ is an increasing function on the nonnegative real numbers, $x^2 > y^2 \implies x > y$. On the other hand, if $y < 0$, then information about $x^2$ and $y^2$ is unneded, since we have immediately that $x > 0 > y$.

Answer 3

Noting that $|x|=x$ for $x\gt 0$, we have$$x^2\gt y^2\Rightarrow |x|\gt|y|\Rightarrow x\gt |y|\ge y\Rightarrow x\gt y.$$

Answer 4

$x^2>y^2\Rightarrow x^2-y^2>0\Rightarrow (x+y)(x-y)>0$, now if $y\geq x$ both factors has to be negative and therefor $x<0$. Hence $x>0\Rightarrow y<x$.