prove inequality $| f(x) - f(y) | < 1/16$











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$f: [ 4, + infty) to mathbb R$



$f(x)= 1/Ax$



Prove: $| f(x) - f(y) | le 1/16 $



I don't know what to do ...

I have no idea... Please help. I have very important test...






analysis inequality







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  • What is the condition on $A$ ?
    – DeepSea
    Nov 25 at 17:18










  • If it is a lipschitz condition ... A = 4
    – Estera Gmiterek
    Nov 25 at 17:21






  • 1




    Is it $1/(Ax)$ or $(1/A)x$?
    – Andrei
    Nov 25 at 17:21















up vote
-4
down vote

favorite












$f: [ 4, + infty) to mathbb R$



$f(x)= 1/Ax$



Prove: $| f(x) - f(y) | le 1/16 $



I don't know what to do ...

I have no idea... Please help. I have very important test...






analysis inequality







share|cite|improve this question
























  • What is the condition on $A$ ?
    – DeepSea
    Nov 25 at 17:18










  • If it is a lipschitz condition ... A = 4
    – Estera Gmiterek
    Nov 25 at 17:21






  • 1




    Is it $1/(Ax)$ or $(1/A)x$?
    – Andrei
    Nov 25 at 17:21













up vote
-4
down vote

favorite









up vote
-4
down vote

favorite











$f: [ 4, + infty) to mathbb R$



$f(x)= 1/Ax$



Prove: $| f(x) - f(y) | le 1/16 $



I don't know what to do ...

I have no idea... Please help. I have very important test...






analysis inequality







share|cite|improve this question















$f: [ 4, + infty) to mathbb R$



$f(x)= 1/Ax$



Prove: $| f(x) - f(y) | le 1/16 $



I don't know what to do ...

I have no idea... Please help. I have very important test...






analysis inequality




analysis inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 17:22









Andrei

10.6k21025




10.6k21025










asked Nov 25 at 17:16









Estera Gmiterek

306




306












  • What is the condition on $A$ ?
    – DeepSea
    Nov 25 at 17:18










  • If it is a lipschitz condition ... A = 4
    – Estera Gmiterek
    Nov 25 at 17:21






  • 1




    Is it $1/(Ax)$ or $(1/A)x$?
    – Andrei
    Nov 25 at 17:21


















  • What is the condition on $A$ ?
    – DeepSea
    Nov 25 at 17:18










  • If it is a lipschitz condition ... A = 4
    – Estera Gmiterek
    Nov 25 at 17:21






  • 1




    Is it $1/(Ax)$ or $(1/A)x$?
    – Andrei
    Nov 25 at 17:21
















What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18




What is the condition on $A$ ?
– DeepSea
Nov 25 at 17:18












If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21




If it is a lipschitz condition ... A = 4
– Estera Gmiterek
Nov 25 at 17:21




1




1




Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21




Is it $1/(Ax)$ or $(1/A)x$?
– Andrei
Nov 25 at 17:21










1 Answer
1






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up vote
4
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Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.






share|cite|improve this answer























  • Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
    – Ingix
    Nov 25 at 20:59











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

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active

oldest

votes








up vote
4
down vote













Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.






share|cite|improve this answer























  • Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
    – Ingix
    Nov 25 at 20:59















up vote
4
down vote













Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.






share|cite|improve this answer























  • Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
    – Ingix
    Nov 25 at 20:59













up vote
4
down vote










up vote
4
down vote









Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.






share|cite|improve this answer














Suppose $A ge 8$, then $|f(x)-f(y)| = left|dfrac{1}{Ax}-dfrac{1}{Ay}right| = dfrac{|x-y|}{Axy} le dfrac{|x|+|y|}{xyA} = dfrac{1}{A}left(dfrac{1}{x}+dfrac{1}{y}right) le dfrac{1}{A}left(dfrac{1}{4}+dfrac{1}{4}right) =dfrac{1}{2A} le dfrac{1}{16}$, as claimed.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 at 17:29

























answered Nov 25 at 17:21









DeepSea

70.7k54487




70.7k54487












  • Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
    – Ingix
    Nov 25 at 20:59


















  • Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
    – Ingix
    Nov 25 at 20:59
















Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59




Can easily be improved to $A ge 4$, since $|f(x)-f(y)| < frac1Amax(frac1x, frac1y)$ as $f$ is positive.
– Ingix
Nov 25 at 20:59


















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