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I've been using this trick a lot lately, I can't seem to prove why it works. E.d. (using an inequality here, but point still remains)

:

$$|x-a| \ge |x| - |a| \tag {1}$$

add $$|x|+|a|\ge |x-a|$$

$$|a|\ge|a|$$

Therefore, $(1)$ is true.

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closed as unclear what you're asking by José Carlos Santos, kingW3, Namaste, Glorfindel, user223391 Jul 18 '17 at 20:55

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    $\begingroup$ You tried to show $\beta\ge \gamma$ from $\alpha\ge \gamma$ and $\alpha\ge \beta$. This won't work, for example if $\alpha=3$, $\beta=1$, $\gamma = 2$. $\endgroup$ – Hagen von Eitzen Jul 15 '17 at 11:29
  • $\begingroup$ @RobertZ aye but I was asking whether it's right to prove things this way. I haven't quite seen this done in any textbook I've read $\endgroup$ – Vrisk Jul 15 '17 at 11:29
  • $\begingroup$ let x=3 ;a=4 then we have you saying: $$|3-4|\ge|3|-|4| $$ aka $$1 \ge -1$$ $$|3|+|4|\ge|3-4|$$ aka $$7\ge 1$$ and $$|4|\gt-|4|$$ true only because the negative sign is outside the absolute value. three nearly unrelated statements. $\endgroup$ – user451844 Jul 15 '17 at 11:34
  • $\begingroup$ a further example would be x=3;a=-1 then we get $$5\ge2$$ $$4\ge5$$ second one clearly false, and we would have gotten ( if not for your edit) $$1\gt-1$$ $\endgroup$ – user451844 Jul 15 '17 at 11:49
  • $\begingroup$ STOP! using it!!!!! It is false and you have been lucky so far that it has worked. All you are doing is proving that something is not mind numbingly obviously false. That does not mean it is true. $\endgroup$ – fleablood Dec 24 '17 at 6:07
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You did

$$\text {unknown inequality}+\text{true inequality} = \text{true inequality}$$

and concluded that the unknown inequality is true. This does not work in general. Take

Unknown inequality: $2>3$

True Inequality: $5>1$

If you add these up you get a true inequality, $7>4$. But as you see this doesn't mean that the unknown inequality must've been true.


Now with equations it's a different matter.

Suppose you have

Unknown equation:$a=b$

True equation: $c=d$

True equation: $a+c=b+d$

To see that the unknown equation is true, just do

$$\text{second true equation} - \text{first true equation}$$

You know that both equations are true so the result of this must be true. But the result is $a=b$, the unknown equation!

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  • $\begingroup$ Oh! Damn I see. $\endgroup$ – Vrisk Jul 15 '17 at 11:44
  • $\begingroup$ I wonder what addition does to the logical validity of the statement, like why does it work for equalities? $\endgroup$ – Vrisk Jul 15 '17 at 11:45
  • $\begingroup$ @Vrisk I edited the answer with that information Hope it helps! $\endgroup$ – Ovi Jul 15 '17 at 11:49
  • $\begingroup$ Oh man, this makes so much sense! Thank you. I now appreciate inverses even more. Like this doesn't work with inequalities because there's no 'inverse' because it's not one to one. Pretty neat $\endgroup$ – Vrisk Jul 15 '17 at 12:00
  • $\begingroup$ @Vrisk Happy to help! $\endgroup$ – Ovi Jul 15 '17 at 12:03
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Just $|x-a|+|a|\geq|x-a+a|=|x|$.

By your reasoning we can prove that $4>5$.

Indeed, since $3>1$ we "obtain": $$3+4>1+5,$$ which is true.

"Hence", $4>5$.

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