Proof $mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$
$begingroup$
We have $(a + b)^{n} geq a^{n} + b^{n}$
Proof:
$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$
binomial-theorem
asked Dec 9 '18 at 19:27
LamineLamine
12
$endgroup$
add a comment |
$begingroup$
We have $(a + b)^{n} geq a^{n} + b^{n}$
Proof:
$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$
binomial-theorem
asked Dec 9 '18 at 19:27
LamineLamine
12
$endgroup$
$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48
$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57
add a comment |
$begingroup$
We have $(a + b)^{n} geq a^{n} + b^{n}$
Proof:
$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$
binomial-theorem
asked Dec 9 '18 at 19:27
LamineLamine
12
$endgroup$
We have $(a + b)^{n} geq a^{n} + b^{n}$
Proof:
$mid a - b mid ^{frac{1}{n}} geq mid mid a mid^{frac{1}{n}} - mid bmid^{frac{1}{n}}mid$
binomial-theorem
binomial-theorem
asked Dec 9 '18 at 19:27
LamineLamine
12
asked Dec 9 '18 at 19:27
LamineLamine
12
edited Dec 9 '18 at 19:36
Lamine
asked Dec 9 '18 at 19:27
LamineLamine
12
asked Dec 9 '18 at 19:27
LamineLamine
12
asked Dec 9 '18 at 19:27
LamineLamine
12
12
$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48
$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57
add a comment |
$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48
$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57
$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48
$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48
$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57
$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can show this by using induction.
https://en.wikipedia.org/wiki/Mathematical_induction#Description
I'll leave it to you to verify it using a base.
Now assume that
$mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)
is true. We want to deduce that
$mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)
is true as well.
Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?
answered Dec 9 '18 at 19:43
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
Raising to the $n$-th power,
$|a - b|^{frac{1}{n}}
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
$
becomes
$|a - b|
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
$.
Letting
$a = c^n$ and $b = d^n$,
this becomes
$|c^n - d^n|
geq || c| - |d||^n
$.
Assuming that
$|c| ge |d|$,
dividing by $c^n$
and letting
$frac{d}{c} = x$,
this is
$|1 - x^n|
geq |1 - x|^n
$.
Assume that $1 ge x ge 0$.
So we want to show that
$1-x^n ge (1-x)^n$.
This is true for $x=0$
and $x=1$.
This is true for $n=1$.
Since $x^n$ and $(1-x)^n$
are decreasing
with increasing $n$,
$1-x^n$ is increasing and
$(1-x)^n$ is decreasing
so the inequality holds
for all $n$.
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can show this by using induction.
https://en.wikipedia.org/wiki/Mathematical_induction#Description
I'll leave it to you to verify it using a base.
Now assume that
$mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)
is true. We want to deduce that
$mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)
is true as well.
Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?
answered Dec 9 '18 at 19:43
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
You can show this by using induction.
https://en.wikipedia.org/wiki/Mathematical_induction#Description
I'll leave it to you to verify it using a base.
Now assume that
$mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)
is true. We want to deduce that
$mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)
is true as well.
Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?
answered Dec 9 '18 at 19:43
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
You can show this by using induction.
https://en.wikipedia.org/wiki/Mathematical_induction#Description
I'll leave it to you to verify it using a base.
Now assume that
$mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)
is true. We want to deduce that
$mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)
is true as well.
Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?
answered Dec 9 '18 at 19:43
CruZCruZ
638
$endgroup$
You can show this by using induction.
https://en.wikipedia.org/wiki/Mathematical_induction#Description
I'll leave it to you to verify it using a base.
Now assume that
$mid(a - b)^{frac{1}{n}}mid leqslant mid a^{frac{1}{n}} - b^{frac{1}{n}}mid$ (1)
is true. We want to deduce that
$mid(a - b)^{frac{1}{n+1}}mid leqslant mid a^{frac{1}{n+1}} - b^{frac{1}{n+1}}mid$ (2)
is true as well.
Hint: Try and see if you can rewrite (2) to (1) in some way and use your assumption. Can you take it from here?
answered Dec 9 '18 at 19:43
CruZCruZ
638
answered Dec 9 '18 at 19:43
CruZCruZ
638
answered Dec 9 '18 at 19:43
CruZCruZ
638
answered Dec 9 '18 at 19:43
CruZCruZ
638
638
add a comment |
add a comment |
$begingroup$
Raising to the $n$-th power,
$|a - b|^{frac{1}{n}}
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
$
becomes
$|a - b|
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
$.
Letting
$a = c^n$ and $b = d^n$,
this becomes
$|c^n - d^n|
geq || c| - |d||^n
$.
Assuming that
$|c| ge |d|$,
dividing by $c^n$
and letting
$frac{d}{c} = x$,
this is
$|1 - x^n|
geq |1 - x|^n
$.
Assume that $1 ge x ge 0$.
So we want to show that
$1-x^n ge (1-x)^n$.
This is true for $x=0$
and $x=1$.
This is true for $n=1$.
Since $x^n$ and $(1-x)^n$
are decreasing
with increasing $n$,
$1-x^n$ is increasing and
$(1-x)^n$ is decreasing
so the inequality holds
for all $n$.
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
$endgroup$
add a comment |
$begingroup$
Raising to the $n$-th power,
$|a - b|^{frac{1}{n}}
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
$
becomes
$|a - b|
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
$.
Letting
$a = c^n$ and $b = d^n$,
this becomes
$|c^n - d^n|
geq || c| - |d||^n
$.
Assuming that
$|c| ge |d|$,
dividing by $c^n$
and letting
$frac{d}{c} = x$,
this is
$|1 - x^n|
geq |1 - x|^n
$.
Assume that $1 ge x ge 0$.
So we want to show that
$1-x^n ge (1-x)^n$.
This is true for $x=0$
and $x=1$.
This is true for $n=1$.
Since $x^n$ and $(1-x)^n$
are decreasing
with increasing $n$,
$1-x^n$ is increasing and
$(1-x)^n$ is decreasing
so the inequality holds
for all $n$.
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
$endgroup$
add a comment |
$begingroup$
Raising to the $n$-th power,
$|a - b|^{frac{1}{n}}
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
$
becomes
$|a - b|
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
$.
Letting
$a = c^n$ and $b = d^n$,
this becomes
$|c^n - d^n|
geq || c| - |d||^n
$.
Assuming that
$|c| ge |d|$,
dividing by $c^n$
and letting
$frac{d}{c} = x$,
this is
$|1 - x^n|
geq |1 - x|^n
$.
Assume that $1 ge x ge 0$.
So we want to show that
$1-x^n ge (1-x)^n$.
This is true for $x=0$
and $x=1$.
This is true for $n=1$.
Since $x^n$ and $(1-x)^n$
are decreasing
with increasing $n$,
$1-x^n$ is increasing and
$(1-x)^n$ is decreasing
so the inequality holds
for all $n$.
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
$endgroup$
Raising to the $n$-th power,
$|a - b|^{frac{1}{n}}
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|
$
becomes
$|a - b|
geq || a|^{frac{1}{n}} - |b|^{frac{1}{n}}|^n
$.
Letting
$a = c^n$ and $b = d^n$,
this becomes
$|c^n - d^n|
geq || c| - |d||^n
$.
Assuming that
$|c| ge |d|$,
dividing by $c^n$
and letting
$frac{d}{c} = x$,
this is
$|1 - x^n|
geq |1 - x|^n
$.
Assume that $1 ge x ge 0$.
So we want to show that
$1-x^n ge (1-x)^n$.
This is true for $x=0$
and $x=1$.
This is true for $n=1$.
Since $x^n$ and $(1-x)^n$
are decreasing
with increasing $n$,
$1-x^n$ is increasing and
$(1-x)^n$ is decreasing
so the inequality holds
for all $n$.
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
answered Dec 9 '18 at 20:49
marty cohenmarty cohen
74.5k549129
74.5k549129
add a comment |
add a comment |
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$begingroup$
What is $a$ and $b$ ? What is $n$ ?
$endgroup$
– Rebellos
Dec 9 '18 at 19:48
$begingroup$
a&b positive real numbers, n natural number*
$endgroup$
– Lamine
Dec 9 '18 at 19:57