Mixing
Generalization of $(a+b)^2leq 2(a^2+b^2)$
$begingroup$
We know that, $(a+b)^2leq 2(a^2+b^2)$. Do we have anything similar for $$left(sum_{i=1}^N a_iright)^2.$$
where $a_iin mathbb{R} forall iin {1,cdots,N}$.
For $n=3$, we get
begin{equation}
begin{aligned}
(a_1+a_2+a_3)^2&leq 2left((a_1+a_2)^2+a_3^2right)
\&leq 2left(2(a_1^2+a_2^2)+a_3^2 right).
end{aligned}
end{equation}
Do we have some sort of generalization?
real-analysis inequality cauchy-schwarz-inequality
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
$endgroup$
add a comment |
$begingroup$
We know that, $(a+b)^2leq 2(a^2+b^2)$. Do we have anything similar for $$left(sum_{i=1}^N a_iright)^2.$$
where $a_iin mathbb{R} forall iin {1,cdots,N}$.
For $n=3$, we get
begin{equation}
begin{aligned}
(a_1+a_2+a_3)^2&leq 2left((a_1+a_2)^2+a_3^2right)
\&leq 2left(2(a_1^2+a_2^2)+a_3^2 right).
end{aligned}
end{equation}
Do we have some sort of generalization?
real-analysis inequality cauchy-schwarz-inequality
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
$endgroup$
1
$begingroup$
Your generalization to $n=3$ is correct, but not symmetric in the variables. A symmetric one would be interesting. Good question.
$endgroup$
– coffeemath
Jan 13 at 22:25
4
$begingroup$
The equality for $n=2$ is equivalent to $0leq (a-b)^2$ so a possible generalizations would be $0 leq (a-b)^2 + (b-c)^2 + (c-a)^2$ and so on (which expanded can be brought on a form similar to what you have). We also have (Cauchy–Schwarz): $(a_1+ldots+a_n)^2 leq n(a_1^2+ldots+a_n^2)$.
$endgroup$
– Winther
Jan 13 at 22:26
add a comment |
$begingroup$
We know that, $(a+b)^2leq 2(a^2+b^2)$. Do we have anything similar for $$left(sum_{i=1}^N a_iright)^2.$$
where $a_iin mathbb{R} forall iin {1,cdots,N}$.
For $n=3$, we get
begin{equation}
begin{aligned}
(a_1+a_2+a_3)^2&leq 2left((a_1+a_2)^2+a_3^2right)
\&leq 2left(2(a_1^2+a_2^2)+a_3^2 right).
end{aligned}
end{equation}
Do we have some sort of generalization?
real-analysis inequality cauchy-schwarz-inequality
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
$endgroup$
We know that, $(a+b)^2leq 2(a^2+b^2)$. Do we have anything similar for $$left(sum_{i=1}^N a_iright)^2.$$
where $a_iin mathbb{R} forall iin {1,cdots,N}$.
For $n=3$, we get
begin{equation}
begin{aligned}
(a_1+a_2+a_3)^2&leq 2left((a_1+a_2)^2+a_3^2right)
\&leq 2left(2(a_1^2+a_2^2)+a_3^2 right).
end{aligned}
end{equation}
Do we have some sort of generalization?
real-analysis inequality cauchy-schwarz-inequality
real-analysis inequality cauchy-schwarz-inequality
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
edited Jan 13 at 23:03
Michael Rozenberg
103k1891195
103k1891195
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
asked Jan 13 at 22:21
Abhinav JhaAbhinav Jha
2741211
2741211
1
$begingroup$
Your generalization to $n=3$ is correct, but not symmetric in the variables. A symmetric one would be interesting. Good question.
$endgroup$
– coffeemath
Jan 13 at 22:25
4
$begingroup$
The equality for $n=2$ is equivalent to $0leq (a-b)^2$ so a possible generalizations would be $0 leq (a-b)^2 + (b-c)^2 + (c-a)^2$ and so on (which expanded can be brought on a form similar to what you have). We also have (Cauchy–Schwarz): $(a_1+ldots+a_n)^2 leq n(a_1^2+ldots+a_n^2)$.
$endgroup$
– Winther
Jan 13 at 22:26
add a comment |
1
$begingroup$
Your generalization to $n=3$ is correct, but not symmetric in the variables. A symmetric one would be interesting. Good question.
$endgroup$
– coffeemath
Jan 13 at 22:25
4
$begingroup$
The equality for $n=2$ is equivalent to $0leq (a-b)^2$ so a possible generalizations would be $0 leq (a-b)^2 + (b-c)^2 + (c-a)^2$ and so on (which expanded can be brought on a form similar to what you have). We also have (Cauchy–Schwarz): $(a_1+ldots+a_n)^2 leq n(a_1^2+ldots+a_n^2)$.
$endgroup$
– Winther
Jan 13 at 22:26
1
1
$begingroup$
Your generalization to $n=3$ is correct, but not symmetric in the variables. A symmetric one would be interesting. Good question.
$endgroup$
– coffeemath
Jan 13 at 22:25
$begingroup$
Your generalization to $n=3$ is correct, but not symmetric in the variables. A symmetric one would be interesting. Good question.
$endgroup$
– coffeemath
Jan 13 at 22:25
4
4
$begingroup$
The equality for $n=2$ is equivalent to $0leq (a-b)^2$ so a possible generalizations would be $0 leq (a-b)^2 + (b-c)^2 + (c-a)^2$ and so on (which expanded can be brought on a form similar to what you have). We also have (Cauchy–Schwarz): $(a_1+ldots+a_n)^2 leq n(a_1^2+ldots+a_n^2)$.
$endgroup$
– Winther
Jan 13 at 22:26
$begingroup$
The equality for $n=2$ is equivalent to $0leq (a-b)^2$ so a possible generalizations would be $0 leq (a-b)^2 + (b-c)^2 + (c-a)^2$ and so on (which expanded can be brought on a form similar to what you have). We also have (Cauchy–Schwarz): $(a_1+ldots+a_n)^2 leq n(a_1^2+ldots+a_n^2)$.
$endgroup$
– Winther
Jan 13 at 22:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)geq(a_1+a_2+...+a_n)^2.$$
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
add a comment |
$begingroup$
Others have mentioned this formula already but I want to remark that we can strengthen it a bit to
$$
|a_1|+cdots+|a_n| le sqrt{n},left(a_1^2+cdots+a_n^2 right)^{1/2}.
$$
Moreover, the "other direction" we also have
$$
left(a_1^2+cdots+a_n^2 right)^{1/2} le |a_1|+cdots+|a_n|,
$$
which follows from super-additivity of $xmapsto x^2$ on $[0,infty)$. Together they show that these two norms on $Bbb R^n$ ($||cdot||_1$ and $||cdot||_2$) are equivalent.
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
The generalization is
$$left(sum_{i=1}^N a_iright)^2le Nsum_{i=1}^N a_i^2$$
which degenerates to equality if all $a_i$ are equal. It is the Cauchy-Schwarz inequality applied to vectors $(1,1,dots,1)$ and $(a_1,a_2,dots,a_n)$.
answered Jan 13 at 22:35
useruser
4,4201929
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)geq(a_1+a_2+...+a_n)^2.$$
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
add a comment |
$begingroup$
It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)geq(a_1+a_2+...+a_n)^2.$$
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
add a comment |
$begingroup$
It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)geq(a_1+a_2+...+a_n)^2.$$
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)geq(a_1+a_2+...+a_n)^2.$$
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 13 at 22:35
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
add a comment |
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
And if the asker is not familiar, C-S is the Cauchy-Schwarz inequality.
$endgroup$
– Theo Bendit
Jan 13 at 22:42
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
So simple! I knew this generalization but wasn't able to find a proof. Thanks for the help!
$endgroup$
– Abhinav Jha
Jan 13 at 22:52
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
$begingroup$
@Abhinav Jha You are welcome!
$endgroup$
– Michael Rozenberg
Jan 13 at 23:03
add a comment |
$begingroup$
Others have mentioned this formula already but I want to remark that we can strengthen it a bit to
$$
|a_1|+cdots+|a_n| le sqrt{n},left(a_1^2+cdots+a_n^2 right)^{1/2}.
$$
Moreover, the "other direction" we also have
$$
left(a_1^2+cdots+a_n^2 right)^{1/2} le |a_1|+cdots+|a_n|,
$$
which follows from super-additivity of $xmapsto x^2$ on $[0,infty)$. Together they show that these two norms on $Bbb R^n$ ($||cdot||_1$ and $||cdot||_2$) are equivalent.
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
Others have mentioned this formula already but I want to remark that we can strengthen it a bit to
$$
|a_1|+cdots+|a_n| le sqrt{n},left(a_1^2+cdots+a_n^2 right)^{1/2}.
$$
Moreover, the "other direction" we also have
$$
left(a_1^2+cdots+a_n^2 right)^{1/2} le |a_1|+cdots+|a_n|,
$$
which follows from super-additivity of $xmapsto x^2$ on $[0,infty)$. Together they show that these two norms on $Bbb R^n$ ($||cdot||_1$ and $||cdot||_2$) are equivalent.
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
$endgroup$
add a comment |
$begingroup$
Others have mentioned this formula already but I want to remark that we can strengthen it a bit to
$$
|a_1|+cdots+|a_n| le sqrt{n},left(a_1^2+cdots+a_n^2 right)^{1/2}.
$$
Moreover, the "other direction" we also have
$$
left(a_1^2+cdots+a_n^2 right)^{1/2} le |a_1|+cdots+|a_n|,
$$
which follows from super-additivity of $xmapsto x^2$ on $[0,infty)$. Together they show that these two norms on $Bbb R^n$ ($||cdot||_1$ and $||cdot||_2$) are equivalent.
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
$endgroup$
Others have mentioned this formula already but I want to remark that we can strengthen it a bit to
$$
|a_1|+cdots+|a_n| le sqrt{n},left(a_1^2+cdots+a_n^2 right)^{1/2}.
$$
Moreover, the "other direction" we also have
$$
left(a_1^2+cdots+a_n^2 right)^{1/2} le |a_1|+cdots+|a_n|,
$$
which follows from super-additivity of $xmapsto x^2$ on $[0,infty)$. Together they show that these two norms on $Bbb R^n$ ($||cdot||_1$ and $||cdot||_2$) are equivalent.
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
answered Jan 13 at 22:57
BigbearZzzBigbearZzz
8,72621652
8,72621652
add a comment |
add a comment |
$begingroup$
The generalization is
$$left(sum_{i=1}^N a_iright)^2le Nsum_{i=1}^N a_i^2$$
which degenerates to equality if all $a_i$ are equal. It is the Cauchy-Schwarz inequality applied to vectors $(1,1,dots,1)$ and $(a_1,a_2,dots,a_n)$.
answered Jan 13 at 22:35
useruser
4,4201929
$endgroup$
add a comment |
$begingroup$
The generalization is
$$left(sum_{i=1}^N a_iright)^2le Nsum_{i=1}^N a_i^2$$
which degenerates to equality if all $a_i$ are equal. It is the Cauchy-Schwarz inequality applied to vectors $(1,1,dots,1)$ and $(a_1,a_2,dots,a_n)$.
answered Jan 13 at 22:35
useruser
4,4201929
$endgroup$
add a comment |
$begingroup$
The generalization is
$$left(sum_{i=1}^N a_iright)^2le Nsum_{i=1}^N a_i^2$$
which degenerates to equality if all $a_i$ are equal. It is the Cauchy-Schwarz inequality applied to vectors $(1,1,dots,1)$ and $(a_1,a_2,dots,a_n)$.
answered Jan 13 at 22:35
useruser
4,4201929
$endgroup$
The generalization is
$$left(sum_{i=1}^N a_iright)^2le Nsum_{i=1}^N a_i^2$$
which degenerates to equality if all $a_i$ are equal. It is the Cauchy-Schwarz inequality applied to vectors $(1,1,dots,1)$ and $(a_1,a_2,dots,a_n)$.
answered Jan 13 at 22:35
useruser
4,4201929
edited Jan 14 at 18:05
answered Jan 13 at 22:35
useruser
4,4201929
answered Jan 13 at 22:35
useruser
4,4201929
answered Jan 13 at 22:35
useruser
4,4201929
4,4201929
add a comment |
add a comment |
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$begingroup$
Your generalization to $n=3$ is correct, but not symmetric in the variables. A symmetric one would be interesting. Good question.
$endgroup$
– coffeemath
Jan 13 at 22:25
4
$begingroup$
The equality for $n=2$ is equivalent to $0leq (a-b)^2$ so a possible generalizations would be $0 leq (a-b)^2 + (b-c)^2 + (c-a)^2$ and so on (which expanded can be brought on a form similar to what you have). We also have (Cauchy–Schwarz): $(a_1+ldots+a_n)^2 leq n(a_1^2+ldots+a_n^2)$.
$endgroup$
– Winther
Jan 13 at 22:26