Prove or disprove $(1-y)^x(1+xy)1$, $0<y<1$.












1














Let$$x>1$$
$$0<y<1$$
Is it possible to prove or disprove this following:
$$(1-y)^x(1+xy)<1$$
I tested many sample results and could not find a case that make it false yet. I think Bernoulli's inequality might help.






calculus inequality







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  • To have symbols inline use single $s either side instead of double.
    – Shaun
    Nov 27 '18 at 2:32






  • 1




    I believe you can use Bernoulli's inequality to get an answer here. Look it up!
    – Gerry Myerson
    Nov 27 '18 at 2:56
















1














Let$$x>1$$
$$0<y<1$$
Is it possible to prove or disprove this following:
$$(1-y)^x(1+xy)<1$$
I tested many sample results and could not find a case that make it false yet. I think Bernoulli's inequality might help.






calculus inequality







share|cite|improve this question
























  • To have symbols inline use single $s either side instead of double.
    – Shaun
    Nov 27 '18 at 2:32






  • 1




    I believe you can use Bernoulli's inequality to get an answer here. Look it up!
    – Gerry Myerson
    Nov 27 '18 at 2:56














1












1








1


1





Let$$x>1$$
$$0<y<1$$
Is it possible to prove or disprove this following:
$$(1-y)^x(1+xy)<1$$
I tested many sample results and could not find a case that make it false yet. I think Bernoulli's inequality might help.






calculus inequality







share|cite|improve this question















Let$$x>1$$
$$0<y<1$$
Is it possible to prove or disprove this following:
$$(1-y)^x(1+xy)<1$$
I tested many sample results and could not find a case that make it false yet. I think Bernoulli's inequality might help.






calculus inequality




calculus inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 2:58

























asked Nov 27 '18 at 2:22









cscisgqr

42




42












  • To have symbols inline use single $s either side instead of double.
    – Shaun
    Nov 27 '18 at 2:32






  • 1




    I believe you can use Bernoulli's inequality to get an answer here. Look it up!
    – Gerry Myerson
    Nov 27 '18 at 2:56


















  • To have symbols inline use single $s either side instead of double.
    – Shaun
    Nov 27 '18 at 2:32






  • 1




    I believe you can use Bernoulli's inequality to get an answer here. Look it up!
    – Gerry Myerson
    Nov 27 '18 at 2:56
















To have symbols inline use single $s either side instead of double.
– Shaun
Nov 27 '18 at 2:32




To have symbols inline use single $s either side instead of double.
– Shaun
Nov 27 '18 at 2:32




1




1




I believe you can use Bernoulli's inequality to get an answer here. Look it up!
– Gerry Myerson
Nov 27 '18 at 2:56




I believe you can use Bernoulli's inequality to get an answer here. Look it up!
– Gerry Myerson
Nov 27 '18 at 2:56










1 Answer
1






active

oldest

votes


















1














Using Bernoulli's Inequality,
$$(1-y)(1+xy)^{1/x}leqslant (1-y)(1+y)=1-y^2<1$$



Raise both sides to the power $x$ to get your desired inequality.






share|cite|improve this answer























  • But this changed my original inequity
    – cscisgqr
    Nov 27 '18 at 3:13






  • 1




    You just take a power of $x$ to both sides to get the desire inequality.
    – Jerry
    Nov 27 '18 at 3:18






  • 1




    @cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
    – Ross Millikan
    Nov 27 '18 at 3:19











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Using Bernoulli's Inequality,
$$(1-y)(1+xy)^{1/x}leqslant (1-y)(1+y)=1-y^2<1$$



Raise both sides to the power $x$ to get your desired inequality.






share|cite|improve this answer























  • But this changed my original inequity
    – cscisgqr
    Nov 27 '18 at 3:13






  • 1




    You just take a power of $x$ to both sides to get the desire inequality.
    – Jerry
    Nov 27 '18 at 3:18






  • 1




    @cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
    – Ross Millikan
    Nov 27 '18 at 3:19
















1














Using Bernoulli's Inequality,
$$(1-y)(1+xy)^{1/x}leqslant (1-y)(1+y)=1-y^2<1$$



Raise both sides to the power $x$ to get your desired inequality.






share|cite|improve this answer























  • But this changed my original inequity
    – cscisgqr
    Nov 27 '18 at 3:13






  • 1




    You just take a power of $x$ to both sides to get the desire inequality.
    – Jerry
    Nov 27 '18 at 3:18






  • 1




    @cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
    – Ross Millikan
    Nov 27 '18 at 3:19














1












1








1






Using Bernoulli's Inequality,
$$(1-y)(1+xy)^{1/x}leqslant (1-y)(1+y)=1-y^2<1$$



Raise both sides to the power $x$ to get your desired inequality.






share|cite|improve this answer














Using Bernoulli's Inequality,
$$(1-y)(1+xy)^{1/x}leqslant (1-y)(1+y)=1-y^2<1$$



Raise both sides to the power $x$ to get your desired inequality.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 3:45

























answered Nov 27 '18 at 3:10









Macavity

35.1k52453




35.1k52453












  • But this changed my original inequity
    – cscisgqr
    Nov 27 '18 at 3:13






  • 1




    You just take a power of $x$ to both sides to get the desire inequality.
    – Jerry
    Nov 27 '18 at 3:18






  • 1




    @cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
    – Ross Millikan
    Nov 27 '18 at 3:19


















  • But this changed my original inequity
    – cscisgqr
    Nov 27 '18 at 3:13






  • 1




    You just take a power of $x$ to both sides to get the desire inequality.
    – Jerry
    Nov 27 '18 at 3:18






  • 1




    @cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
    – Ross Millikan
    Nov 27 '18 at 3:19
















But this changed my original inequity
– cscisgqr
Nov 27 '18 at 3:13




But this changed my original inequity
– cscisgqr
Nov 27 '18 at 3:13




1




1




You just take a power of $x$ to both sides to get the desire inequality.
– Jerry
Nov 27 '18 at 3:18




You just take a power of $x$ to both sides to get the desire inequality.
– Jerry
Nov 27 '18 at 3:18




1




1




@cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
– Ross Millikan
Nov 27 '18 at 3:19




@cscisgqr: It is just the $x$ root of your original inequality. Since we are comparing with $1$, that works because $1^{1/x}=1$
– Ross Millikan
Nov 27 '18 at 3:19


















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