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So the title says it all. I think using the square formula, not sure if that is legit. Thank you for your attention.

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    $\begingroup$ It is legit. Just square both sides in the first equation, and you get the answer. You will need to use $y^2\ge 0$. $\endgroup$ – Andrei Apr 2 at 17:59
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The inequality $$|{x-0}| > |{x-y}|$$ says that $x$ is farther away from zero than it is from $y$, so both $x$ and $y$ are on the same side of zero, meaning they have the same sign.

Thus $xy > 0$.

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Let $|x|>|x-y|$. Then $x^2>(x-y)^2=x^2-2xy+y^2$ but $y^2\ge 0$, so $0>-2xy$, which means $xy>0$ as inequalities are reversed when multiplying by a negative.

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In general, $a>b$ is not equivalent to $a^2>b^2.$ However, when $a,b\ge 0,$ we can say that $a^2>b^2$ if and only if $a>b.$

On the one hand, suppose $a>b$ and $a,b\ge 0.$ Then $a>0,$ so since $a>b,$ then $a\cdot a>a\cdot b,$ or $a^2>ab.$ Since $b\ge 0$ and $a>b,$ then $b\cdot a\ge b\cdot b,$ or $ab\ge b^2.$ Since $a^2>ab\ge b^2,$ then $a^2>b^2.$

On the other hand, suppose $a^2>b^2$ and $a,b\ge 0.$ Then $a^2-b^2>0,$ or $(a-b)(a+b)>0,$ meaning that $a-b$ and $a+b$ are either both positive or both negative. Since $a,b\ge 0,$ then $a+b$ can't be negative, and so both $a-b$ and $a+b$ are positive. Then $a-b>0,$ and so $a>b.$


The upshot is that your approach absolutely works in this case, but we might want to approach it differently. Personally, I'd think about it in terms of distances (as in Alice's answer).

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Taking cue from Alice and Shaun,

Since $|x|$ and $|x-y|\geq0$, $y\neq0$ with $|x|>|x-y|$ so $$x^2>(x-y)^2\implies0>y(y-2x)$$ This happens iff only one of $y$ or $y-2x$ is less than $0$.

  • If $y<0$ then $y-2x>0\implies y>2x\implies x<0$

  • If $y>0$ then $y-2x<0\implies y<2x\implies x>0$

In either case $x$ and $y$ have same sign so $xy>0$.

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