Help applying the triangle inequality

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I've been stuck on this problem for a while, not sure how to tackle it.



$$textLet a,b in mathbbR. textIf 0 < epsilon < textmina textshow that$$
$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilon$$



You're just meant to apply the triangle inequality but I'm not sure how to split up the LHS or how to get the negative on the RHS. Also not sure where to use $e < mina$ :/



Any help is appreciated.






real-analysis inequality







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    Hint: note that $left|frac xyright| = fracy$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality.
    – Mees de Vries
    Sep 10 at 12:44














up vote
0
down vote

favorite












I've been stuck on this problem for a while, not sure how to tackle it.



$$textLet a,b in mathbbR. textIf 0 < epsilon < textmina textshow that$$
$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilon$$



You're just meant to apply the triangle inequality but I'm not sure how to split up the LHS or how to get the negative on the RHS. Also not sure where to use $e < mina$ :/



Any help is appreciated.






real-analysis inequality







share|cite|improve this question

















  • 2




    Hint: note that $left|frac xyright| = fracy$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality.
    – Mees de Vries
    Sep 10 at 12:44












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I've been stuck on this problem for a while, not sure how to tackle it.



$$textLet a,b in mathbbR. textIf 0 < epsilon < textmina textshow that$$
$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilon$$



You're just meant to apply the triangle inequality but I'm not sure how to split up the LHS or how to get the negative on the RHS. Also not sure where to use $e < mina$ :/



Any help is appreciated.






real-analysis inequality







share|cite|improve this question













I've been stuck on this problem for a while, not sure how to tackle it.



$$textLet a,b in mathbbR. textIf 0 < epsilon < textmina textshow that$$
$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilon$$



You're just meant to apply the triangle inequality but I'm not sure how to split up the LHS or how to get the negative on the RHS. Also not sure where to use $e < mina$ :/



Any help is appreciated.






real-analysis inequality




real-analysis inequality






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share|cite|improve this question











share|cite|improve this question




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asked Sep 10 at 12:42









tzcl

82




82







  • 2




    Hint: note that $left|frac xyright| = fracy$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality.
    – Mees de Vries
    Sep 10 at 12:44












  • 2




    Hint: note that $left|frac xyright| = fracy$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality.
    – Mees de Vries
    Sep 10 at 12:44







2




2




Hint: note that $left|frac xyright| = fracy$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality.
– Mees de Vries
Sep 10 at 12:44




Hint: note that $left|frac xyright| = fracy$, and then note that each denominator is positive so you can cross multiply and get an equivalent inequality.
– Mees de Vries
Sep 10 at 12:44










1 Answer
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Fact 1: If $a,b$ are real numbers and $a<b$ and $c>0$ then $ac<bc$



Now by triangle inequality



$|a+epsilon|leq |a|+|epsilon|=|a|+epsilon$



as $epsilon$ is positive number.



So, using fact $1$,



$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilonb+ epsilon verttag1label eq1$$



Now using the reverse triangle inequality $|a+b|geq||a|-|b||$



$|b+epsilon|geq ||b|-epsilon|=|b|-epsilon$



This follows from the condition given on $epsilon$



So $$frac1leq frac1b$$



Again using the fact 1,



$$frac + epsilonb+ epsilon vertleq frac + epsilonbtag2 $$



Using $(1)$ and $(2)$, you get the desired inequality.






share|cite|improve this answer
















  • 1




    Thanks so much, didn't think of doing it like this
    – tzcl
    Sep 11 at 1:07










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Fact 1: If $a,b$ are real numbers and $a<b$ and $c>0$ then $ac<bc$



Now by triangle inequality



$|a+epsilon|leq |a|+|epsilon|=|a|+epsilon$



as $epsilon$ is positive number.



So, using fact $1$,



$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilonb+ epsilon verttag1label eq1$$



Now using the reverse triangle inequality $|a+b|geq||a|-|b||$



$|b+epsilon|geq ||b|-epsilon|=|b|-epsilon$



This follows from the condition given on $epsilon$



So $$frac1leq frac1b$$



Again using the fact 1,



$$frac + epsilonb+ epsilon vertleq frac + epsilonbtag2 $$



Using $(1)$ and $(2)$, you get the desired inequality.






share|cite|improve this answer
















  • 1




    Thanks so much, didn't think of doing it like this
    – tzcl
    Sep 11 at 1:07














up vote
1
down vote



accepted










Fact 1: If $a,b$ are real numbers and $a<b$ and $c>0$ then $ac<bc$



Now by triangle inequality



$|a+epsilon|leq |a|+|epsilon|=|a|+epsilon$



as $epsilon$ is positive number.



So, using fact $1$,



$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilonb+ epsilon verttag1label eq1$$



Now using the reverse triangle inequality $|a+b|geq||a|-|b||$



$|b+epsilon|geq ||b|-epsilon|=|b|-epsilon$



This follows from the condition given on $epsilon$



So $$frac1leq frac1b$$



Again using the fact 1,



$$frac + epsilonb+ epsilon vertleq frac + epsilonbtag2 $$



Using $(1)$ and $(2)$, you get the desired inequality.






share|cite|improve this answer
















  • 1




    Thanks so much, didn't think of doing it like this
    – tzcl
    Sep 11 at 1:07












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Fact 1: If $a,b$ are real numbers and $a<b$ and $c>0$ then $ac<bc$



Now by triangle inequality



$|a+epsilon|leq |a|+|epsilon|=|a|+epsilon$



as $epsilon$ is positive number.



So, using fact $1$,



$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilonb+ epsilon verttag1label eq1$$



Now using the reverse triangle inequality $|a+b|geq||a|-|b||$



$|b+epsilon|geq ||b|-epsilon|=|b|-epsilon$



This follows from the condition given on $epsilon$



So $$frac1leq frac1b$$



Again using the fact 1,



$$frac + epsilonb+ epsilon vertleq frac + epsilonbtag2 $$



Using $(1)$ and $(2)$, you get the desired inequality.






share|cite|improve this answer












Fact 1: If $a,b$ are real numbers and $a<b$ and $c>0$ then $ac<bc$



Now by triangle inequality



$|a+epsilon|leq |a|+|epsilon|=|a|+epsilon$



as $epsilon$ is positive number.



So, using fact $1$,



$$left vert fraca + epsilonb + epsilon right vert leq frac + epsilonb+ epsilon verttag1label eq1$$



Now using the reverse triangle inequality $|a+b|geq||a|-|b||$



$|b+epsilon|geq ||b|-epsilon|=|b|-epsilon$



This follows from the condition given on $epsilon$



So $$frac1leq frac1b$$



Again using the fact 1,



$$frac + epsilonb+ epsilon vertleq frac + epsilonbtag2 $$



Using $(1)$ and $(2)$, you get the desired inequality.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 10 at 17:26









StammeringMathematician

86017




86017







  • 1




    Thanks so much, didn't think of doing it like this
    – tzcl
    Sep 11 at 1:07












  • 1




    Thanks so much, didn't think of doing it like this
    – tzcl
    Sep 11 at 1:07







1




1




Thanks so much, didn't think of doing it like this
– tzcl
Sep 11 at 1:07




Thanks so much, didn't think of doing it like this
– tzcl
Sep 11 at 1:07

















 

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