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I was doing the following question and wonder if my proof below is valid

Prove that $\frac{x+z}{y+z} > \frac{x}{y} \to x<y$ if and only if $x<y$

To prove that $\frac{x+z}{y+z} > \frac{x}{y} \to x<y$

$$(x+z)y > (y+z)x$$

$$xy+zy > xy + xz$$

$$zy > xz$$

$$y>x$$

Was I allowed to divide both sides by $z$?

And to prove the other direction, am I able to just write down the previous proof in the reverse direction? (i.e. start with $y>x$

Any hints wouldbe appreciated!

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  • $\begingroup$ Is $z$ supposed to be positive? $\endgroup$
    – dmk
    Commented Dec 22, 2022 at 0:56
  • $\begingroup$ Welcome. Your proof doesn’t work if either of $z$ or $y+z$ or $y$ are negative, because then you’re flipping the inequality sign when you cross multiply. These effects balance out if you’re more careful, though $\endgroup$
    – FShrike
    Commented Dec 22, 2022 at 0:56
  • $\begingroup$ Yeah that's what I was afraid of unfortunately. Would my proof work if x, y, z > 0? $\endgroup$
    – austiyi12345
    Commented Dec 22, 2022 at 1:07
  • $\begingroup$ Yes, except $x$ is always in the top so no need there. You multiplied by $y$ and by $y+z$ at the beginning and divided by $z$ toward the end. You need those to be positive to keep the direction of the inequality. More generally, if $y,w>0$ and if $x/y\le z/w$, then $x/y\le (x+z)/(y+w)\le z/w$ with equality iff $x/y=z/w$. Check out my all time favorite Numberphile, youtube.com/watch?v=0hlvhQZIOQw. $\endgroup$
    – Aaron Goldsmith
    Commented Dec 22, 2022 at 1:37
  • $\begingroup$ Great, thanks a lot for the help everyone! $\endgroup$
    – austiyi12345
    Commented Dec 22, 2022 at 2:23

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