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anonymous
 4 years ago
im trying to prove the inverse triangle inequality\[xy\leqxy\]like this:\[x=xy+y,\]\[x=xy+y,\]\[x\leqxy+y,\]\[xy\leqxy.\]i get stuck here; dont know if its legit or not to take the absolute value of the LHS and preserve the inequality. help?
anonymous
 4 years ago
im trying to prove the inverse triangle inequality\[xy\leqxy\]like this:\[x=xy+y,\]\[x=xy+y,\]\[x\leqxy+y,\]\[xy\leqxy.\]i get stuck here; dont know if its legit or not to take the absolute value of the LHS and preserve the inequality. help?

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watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2your statement is equivalent to \(xy\leq xy\leq xy\) now use triangle inequality

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im a little confused... how can i do that?

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2Do you know that the inequality \(x\leq a\) is equivalent to \(a\leq x\leq a\) ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i agree, but how can i apply the triangle inequality to the above system?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i thought i could only apply it to an expression.

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2ok, let's look at the first inequality \(xy\leq xy\). Can you rewrite the inequality so that there is no minus (but all plus) by moving around the expressions?

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2Can you see that it is equivalent to \(y\leq x+xy\)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[xy\leqxy\impliesy\leqx+xy?\]

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2Now if we apply the triangle inequality to \(y=(yx)+x\) what do we get?

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2yes, but that is the same as \(y\leq x+xy\) right ? since yx=xy

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2so if you want a more linear argument we can like this for the first inequality: \(y=(yx)+x\leq yx+x=x+xy\) It follows that \(xy\leq xy\)

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2you can do something similar to the 2nd one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that makes sense. and with this, we can conclude that\[xy\leqxy?\]

watchmath
 4 years ago
Best ResponseYou've already chosen the best response.2yes, if you prove two inequalities: \(xy\leq xy\) and \(xy\leq xy\) you conclude the original inequality

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0beautiful. i see it now. thank you very much for your time and help. i appreciate it a lot!
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