Proof of the inequality $x^p - x^p+1 leq frac1p+1$ for $p in [0,1]$ [closed]

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Let $x in [0,1]$. How can I show the inequality $x^p - x^p+1 leq frac1p+1$?




calculus real-analysis analysis inequality




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closed as off-topic by José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy Jul 21 at 15:53


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Let $x in [0,1]$. How can I show the inequality $x^p - x^p+1 leq frac1p+1$?




calculus real-analysis analysis inequality




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closed as off-topic by José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy Jul 21 at 15:53


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." – José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.












  • This should be so simple you would already have some thoughts in your head when you penned this question down.
    – Parcly Taxel
    Jul 21 at 15:04












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Let $x in [0,1]$. How can I show the inequality $x^p - x^p+1 leq frac1p+1$?




calculus real-analysis analysis inequality




share|cite|improve this question













Let $x in [0,1]$. How can I show the inequality $x^p - x^p+1 leq frac1p+1$?





calculus real-analysis analysis inequality





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edited Jul 21 at 15:15









Jneven

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asked Jul 21 at 15:01









user144921

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closed as off-topic by José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy Jul 21 at 15:53


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." – José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy Jul 21 at 15:53


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." – José Carlos Santos, Siong Thye Goh, Shailesh, Adrian Keister, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.











  • This should be so simple you would already have some thoughts in your head when you penned this question down.
    – Parcly Taxel
    Jul 21 at 15:04
















  • This should be so simple you would already have some thoughts in your head when you penned this question down.
    – Parcly Taxel
    Jul 21 at 15:04















This should be so simple you would already have some thoughts in your head when you penned this question down.
– Parcly Taxel
Jul 21 at 15:04




This should be so simple you would already have some thoughts in your head when you penned this question down.
– Parcly Taxel
Jul 21 at 15:04










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Hint: We have by the Weighted AM-GM Inequality that $$x^p(p-px)leq left(fracpx+(p-px)p+1right)^p+1leq fracpp+1,.$$
(Here, $p$ can be any positive real number.)






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    1 Answer
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    Hint: We have by the Weighted AM-GM Inequality that $$x^p(p-px)leq left(fracpx+(p-px)p+1right)^p+1leq fracpp+1,.$$
    (Here, $p$ can be any positive real number.)






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      Hint: We have by the Weighted AM-GM Inequality that $$x^p(p-px)leq left(fracpx+(p-px)p+1right)^p+1leq fracpp+1,.$$
      (Here, $p$ can be any positive real number.)






      share|cite|improve this answer























        up vote
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        up vote
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        Hint: We have by the Weighted AM-GM Inequality that $$x^p(p-px)leq left(fracpx+(p-px)p+1right)^p+1leq fracpp+1,.$$
        (Here, $p$ can be any positive real number.)






        share|cite|improve this answer













        Hint: We have by the Weighted AM-GM Inequality that $$x^p(p-px)leq left(fracpx+(p-px)p+1right)^p+1leq fracpp+1,.$$
        (Here, $p$ can be any positive real number.)







        share|cite|improve this answer













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        answered Jul 21 at 15:12









        Batominovski

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