Question on a inequality satisfied by real numbers [on hold]

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For two reals $a,b$ why the inequality $|a-b|^pleq 2^p maxa$ true for $1<p<infty$?




inequality




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put on hold as off-topic by Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah Aug 3 at 14:15


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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah
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  • Without loss of generality, let $|a| geq |b|$ and divide both sides by $|a|^p$.
    – Brian Tung
    Aug 3 at 6:04














up vote
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For two reals $a,b$ why the inequality $|a-b|^pleq 2^p maxa$ true for $1<p<infty$?




inequality




share|cite|improve this question











put on hold as off-topic by Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah Aug 3 at 14:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Without loss of generality, let $|a| geq |b|$ and divide both sides by $|a|^p$.
    – Brian Tung
    Aug 3 at 6:04












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











For two reals $a,b$ why the inequality $|a-b|^pleq 2^p maxa$ true for $1<p<infty$?




inequality




share|cite|improve this question











For two reals $a,b$ why the inequality $|a-b|^pleq 2^p maxa$ true for $1<p<infty$?





inequality





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 3 at 5:59









Tanmoy Paul

1




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put on hold as off-topic by Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah Aug 3 at 14:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah Aug 3 at 14:15


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Jose Arnaldo Bebita Dris, Arnaud D., Claude Leibovici, Micah
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Without loss of generality, let $|a| geq |b|$ and divide both sides by $|a|^p$.
    – Brian Tung
    Aug 3 at 6:04
















  • Without loss of generality, let $|a| geq |b|$ and divide both sides by $|a|^p$.
    – Brian Tung
    Aug 3 at 6:04















Without loss of generality, let $|a| geq |b|$ and divide both sides by $|a|^p$.
– Brian Tung
Aug 3 at 6:04




Without loss of generality, let $|a| geq |b|$ and divide both sides by $|a|^p$.
– Brian Tung
Aug 3 at 6:04










1 Answer
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WLOG: $|a| le |b|$. Then $maxa=|b|^p$. We then have



$|a-b| le |a|+|b| le 2|b|$, thus



$$|a-b|^p le 2^p|b|^p.$$






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  • Why the downvote ???????????????????
    – Fred
    yesterday

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
-1
down vote













WLOG: $|a| le |b|$. Then $maxa=|b|^p$. We then have



$|a-b| le |a|+|b| le 2|b|$, thus



$$|a-b|^p le 2^p|b|^p.$$






share|cite|improve this answer





















  • Why the downvote ???????????????????
    – Fred
    yesterday














up vote
-1
down vote













WLOG: $|a| le |b|$. Then $maxa=|b|^p$. We then have



$|a-b| le |a|+|b| le 2|b|$, thus



$$|a-b|^p le 2^p|b|^p.$$






share|cite|improve this answer





















  • Why the downvote ???????????????????
    – Fred
    yesterday












up vote
-1
down vote










up vote
-1
down vote









WLOG: $|a| le |b|$. Then $maxa=|b|^p$. We then have



$|a-b| le |a|+|b| le 2|b|$, thus



$$|a-b|^p le 2^p|b|^p.$$






share|cite|improve this answer













WLOG: $|a| le |b|$. Then $maxa=|b|^p$. We then have



$|a-b| le |a|+|b| le 2|b|$, thus



$$|a-b|^p le 2^p|b|^p.$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 3 at 6:08









Fred

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  • Why the downvote ???????????????????
    – Fred
    yesterday
















  • Why the downvote ???????????????????
    – Fred
    yesterday















Why the downvote ???????????????????
– Fred
yesterday




Why the downvote ???????????????????
– Fred
yesterday


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