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This question already has an answer here:

Suppose $x,y$ are both positive real numbers. Prove that $(x^2 - y^2)(1/y - 1/x) \geq 0$

I've tried several ways but still stuck on this one, please give me some ideas, thanks a lot.

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marked as duplicate by dxiv, Community May 13 '17 at 3:04

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Show that $(x^2 -y^2)(1/y - 1/x)$ = $\frac{(x^2 - y^2)(x -y)}{xy} \ge 0$ for $ x,y \gt 0$.

Denominator $xy$ is positive, multiply inequality by $xy$ to get:

$(x^2 - y^2)(x -y) \ge 0$.

1) Let $x \ge y$ then:

$(x^2 - y^2)(x - y)$ = $(x- y)^2(x +y) \ge 0$,

a square, $(x -y)^2$, multiplied by a pos. number $(x + y)$.

2) Let $y \gt x$ then:

$(x^2 - y^2)(x-y)$ =

$(y^2 - x^2)(y -x) $ =

$ (y - x)^2(y +x)$,

same argument as in 1).

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