Prove or disprove $(1-y)^x(1+xy)1$, $0<y<1$.
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
asked Nov 27 '18 at 2:22
cscisgqr
42
add a comment |
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
asked Nov 27 '18 at 2:22
cscisgqr
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
add a comment |
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
asked Nov 27 '18 at 2:22
cscisgqr
42
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
asked Nov 27 '18 at 2:22
cscisgqr
42
asked Nov 27 '18 at 2:22
cscisgqr
42
edited Nov 27 '18 at 2:58
asked Nov 27 '18 at 2:22
cscisgqr
42
asked Nov 27 '18 at 2:22
cscisgqr
42
asked Nov 27 '18 at 2:22
cscisgqr
42
42
To have symbols inline use single$s either side instead of double.
– Shaun
Nov 27 '18 at 2:32
1
I believe you can use Bernoulli's inequality to get an answer here. Look it up!
– Gerry Myerson
Nov 27 '18 at 2:56
add a comment |
To have symbols inline use single$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
add a comment |
1 Answer
1
active
oldest
votes
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.
answered Nov 27 '18 at 3:10
Macavity
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
add a comment |
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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
oldest
votes
active
oldest
votes
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.
answered Nov 27 '18 at 3:10
Macavity
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
add a comment |
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.
answered Nov 27 '18 at 3:10
Macavity
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
add a comment |
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.
answered Nov 27 '18 at 3:10
Macavity
35.1k52453
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.
answered Nov 27 '18 at 3:10
Macavity
35.1k52453
edited Nov 27 '18 at 3:45
answered Nov 27 '18 at 3:10
Macavity
35.1k52453
answered Nov 27 '18 at 3:10
Macavity
35.1k52453
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
add a comment |
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
add a comment |
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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