Proving the following quadratic inequality? [duplicate]
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Apologies if this has been asked before - I could not find a question with this exact inequality.
Basically the inequality is
$$(a+b+c)^2 leq 3 a^2 + 3 b^2 + 3 c^2$$
Expanding it out we see that
$$(a+b+c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ac$$
so I guess it is equivalent to showing that
$$ab + bc + ac leq a^2 + b^2 + c^2$$
Which makes sense to me. But how exactly do I prove it?
We can assume WLOG that each $a,b,c > 0$ since $ab leq |a||b|$. From here, I guess we need to show that
$$ab leq frac{1}{2} left(max(a,b)^2 + min(a,b)^2
right)$$
And the result follows by adding up each term. But I'm not really sure why this must hold.
inequality self-learning
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
marked as duplicate by Martin R, Cyclohexanol., stressed out, user10354138, Joel Reyes Noche Dec 19 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Apologies if this has been asked before - I could not find a question with this exact inequality.
Basically the inequality is
$$(a+b+c)^2 leq 3 a^2 + 3 b^2 + 3 c^2$$
Expanding it out we see that
$$(a+b+c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ac$$
so I guess it is equivalent to showing that
$$ab + bc + ac leq a^2 + b^2 + c^2$$
Which makes sense to me. But how exactly do I prove it?
We can assume WLOG that each $a,b,c > 0$ since $ab leq |a||b|$. From here, I guess we need to show that
$$ab leq frac{1}{2} left(max(a,b)^2 + min(a,b)^2
right)$$
And the result follows by adding up each term. But I'm not really sure why this must hold.
inequality self-learning
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
marked as duplicate by Martin R, Cyclohexanol., stressed out, user10354138, Joel Reyes Noche Dec 19 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Yes we will prove $a^2+b^2+c^2ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
– Word Shallow
Dec 19 '18 at 8:22
A more general case including your one can be found here: math.stackexchange.com/questions/1249986/prove-qm-am-inequality
– trancelocation
Dec 19 '18 at 8:25
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 leq 3(a^2+b^2+c^2)$
– Martin R
Dec 19 '18 at 8:33
add a comment |
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Apologies if this has been asked before - I could not find a question with this exact inequality.
Basically the inequality is
$$(a+b+c)^2 leq 3 a^2 + 3 b^2 + 3 c^2$$
Expanding it out we see that
$$(a+b+c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ac$$
so I guess it is equivalent to showing that
$$ab + bc + ac leq a^2 + b^2 + c^2$$
Which makes sense to me. But how exactly do I prove it?
We can assume WLOG that each $a,b,c > 0$ since $ab leq |a||b|$. From here, I guess we need to show that
$$ab leq frac{1}{2} left(max(a,b)^2 + min(a,b)^2
right)$$
And the result follows by adding up each term. But I'm not really sure why this must hold.
inequality self-learning
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
Apologies if this has been asked before - I could not find a question with this exact inequality.
Basically the inequality is
$$(a+b+c)^2 leq 3 a^2 + 3 b^2 + 3 c^2$$
Expanding it out we see that
$$(a+b+c)^2 = a^2 +b^2 + c^2 + 2ab + 2bc + 2ac$$
so I guess it is equivalent to showing that
$$ab + bc + ac leq a^2 + b^2 + c^2$$
Which makes sense to me. But how exactly do I prove it?
We can assume WLOG that each $a,b,c > 0$ since $ab leq |a||b|$. From here, I guess we need to show that
$$ab leq frac{1}{2} left(max(a,b)^2 + min(a,b)^2
right)$$
And the result follows by adding up each term. But I'm not really sure why this must hold.
This question already has an answer here:
How to prove $a^2 + b^2 + c^2 ge ab + bc + ca$?
6 answers
inequality self-learning
inequality self-learning
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
asked Dec 19 '18 at 8:18
Xiaomi
1,016115
1,016115
marked as duplicate by Martin R, Cyclohexanol., stressed out, user10354138, Joel Reyes Noche Dec 19 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Cyclohexanol., stressed out, user10354138, Joel Reyes Noche Dec 19 '18 at 13:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
2
Yes we will prove $a^2+b^2+c^2ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
– Word Shallow
Dec 19 '18 at 8:22
A more general case including your one can be found here: math.stackexchange.com/questions/1249986/prove-qm-am-inequality
– trancelocation
Dec 19 '18 at 8:25
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 leq 3(a^2+b^2+c^2)$
– Martin R
Dec 19 '18 at 8:33
add a comment |
2
Yes we will prove $a^2+b^2+c^2ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
– Word Shallow
Dec 19 '18 at 8:22
A more general case including your one can be found here: math.stackexchange.com/questions/1249986/prove-qm-am-inequality
– trancelocation
Dec 19 '18 at 8:25
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 leq 3(a^2+b^2+c^2)$
– Martin R
Dec 19 '18 at 8:33
2
2
Yes we will prove $a^2+b^2+c^2ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
– Word Shallow
Dec 19 '18 at 8:22
Yes we will prove $a^2+b^2+c^2ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
– Word Shallow
Dec 19 '18 at 8:22
A more general case including your one can be found here: math.stackexchange.com/questions/1249986/prove-qm-am-inequality
– trancelocation
Dec 19 '18 at 8:25
A more general case including your one can be found here: math.stackexchange.com/questions/1249986/prove-qm-am-inequality
– trancelocation
Dec 19 '18 at 8:25
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 leq 3(a^2+b^2+c^2)$
– Martin R
Dec 19 '18 at 8:33
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 leq 3(a^2+b^2+c^2)$
– Martin R
Dec 19 '18 at 8:33
add a comment |
4 Answers
4
active
oldest
votes
Your last step follows easily from $(a-b)^{2} geq 0$. (Consider the cases $a<b$ and $a geq b$).
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
add a comment |
It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1cdot a + 1 cdot b + 1 cdot c)^2leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
add a comment |
Once you get your final inequality you simply multiply both sides by $2$ and you get $$2a^2 + 2b^2 + 2c^2 - 2ab -2bc-2ca geq 0$$ This can be rewritten as $$a^2 + b^2 - 2ab + a^2 + c^2 - 2ac + b^2 + c^2 -2cageq 0$$ Which translates to $$Biggl(Bigl(a-bbigl)^2 + Bigl(b-cBigl)^2 + Bigl(c-abigl)^2Biggl) geq 0$$ Which is true $forall ; a,b,c ; epsilon ; mathbb{R}.$
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
add a comment |
It is $$(a-b)^2+(b-c)^2+(c-a)^2geq 0$$
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your last step follows easily from $(a-b)^{2} geq 0$. (Consider the cases $a<b$ and $a geq b$).
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
add a comment |
Your last step follows easily from $(a-b)^{2} geq 0$. (Consider the cases $a<b$ and $a geq b$).
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
add a comment |
Your last step follows easily from $(a-b)^{2} geq 0$. (Consider the cases $a<b$ and $a geq b$).
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
Your last step follows easily from $(a-b)^{2} geq 0$. (Consider the cases $a<b$ and $a geq b$).
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
answered Dec 19 '18 at 8:21
Kavi Rama Murthy
50.3k31854
50.3k31854
add a comment |
add a comment |
It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1cdot a + 1 cdot b + 1 cdot c)^2leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
add a comment |
It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1cdot a + 1 cdot b + 1 cdot c)^2leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
add a comment |
It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1cdot a + 1 cdot b + 1 cdot c)^2leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
It follows immediately from Cauchy-Schwarz:
$$(a+b+c)^2 = (1cdot a + 1 cdot b + 1 cdot c)^2leq (1^2+1^2+1^2) (a^2 + b^2 + c^2)$$
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
answered Dec 19 '18 at 8:29
trancelocation
9,1051521
9,1051521
add a comment |
add a comment |
Once you get your final inequality you simply multiply both sides by $2$ and you get $$2a^2 + 2b^2 + 2c^2 - 2ab -2bc-2ca geq 0$$ This can be rewritten as $$a^2 + b^2 - 2ab + a^2 + c^2 - 2ac + b^2 + c^2 -2cageq 0$$ Which translates to $$Biggl(Bigl(a-bbigl)^2 + Bigl(b-cBigl)^2 + Bigl(c-abigl)^2Biggl) geq 0$$ Which is true $forall ; a,b,c ; epsilon ; mathbb{R}.$
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
add a comment |
Once you get your final inequality you simply multiply both sides by $2$ and you get $$2a^2 + 2b^2 + 2c^2 - 2ab -2bc-2ca geq 0$$ This can be rewritten as $$a^2 + b^2 - 2ab + a^2 + c^2 - 2ac + b^2 + c^2 -2cageq 0$$ Which translates to $$Biggl(Bigl(a-bbigl)^2 + Bigl(b-cBigl)^2 + Bigl(c-abigl)^2Biggl) geq 0$$ Which is true $forall ; a,b,c ; epsilon ; mathbb{R}.$
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
add a comment |
Once you get your final inequality you simply multiply both sides by $2$ and you get $$2a^2 + 2b^2 + 2c^2 - 2ab -2bc-2ca geq 0$$ This can be rewritten as $$a^2 + b^2 - 2ab + a^2 + c^2 - 2ac + b^2 + c^2 -2cageq 0$$ Which translates to $$Biggl(Bigl(a-bbigl)^2 + Bigl(b-cBigl)^2 + Bigl(c-abigl)^2Biggl) geq 0$$ Which is true $forall ; a,b,c ; epsilon ; mathbb{R}.$
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
Once you get your final inequality you simply multiply both sides by $2$ and you get $$2a^2 + 2b^2 + 2c^2 - 2ab -2bc-2ca geq 0$$ This can be rewritten as $$a^2 + b^2 - 2ab + a^2 + c^2 - 2ac + b^2 + c^2 -2cageq 0$$ Which translates to $$Biggl(Bigl(a-bbigl)^2 + Bigl(b-cBigl)^2 + Bigl(c-abigl)^2Biggl) geq 0$$ Which is true $forall ; a,b,c ; epsilon ; mathbb{R}.$
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
answered Dec 19 '18 at 8:27
Prakhar Nagpal
724318
724318
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
add a comment |
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
And in fact multiplying by $2$ is only necessary because of having divided by $2$ previously! I went straight through $2ab+2bc+2acle 2a^2+2b^2+2c^2$ when I attempted it, and totally missed the massive clue in $-2ab$. I saw all those $2$'s and instinctively divided.
– timtfj
Dec 19 '18 at 16:46
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
yes I was going to mention why he divided the $2$ but then I just decided to leave it
– Prakhar Nagpal
Dec 19 '18 at 18:30
add a comment |
It is $$(a-b)^2+(b-c)^2+(c-a)^2geq 0$$
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
add a comment |
It is $$(a-b)^2+(b-c)^2+(c-a)^2geq 0$$
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
add a comment |
It is $$(a-b)^2+(b-c)^2+(c-a)^2geq 0$$
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
It is $$(a-b)^2+(b-c)^2+(c-a)^2geq 0$$
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
answered Dec 19 '18 at 8:22
Dr. Sonnhard Graubner
73k42865
73k42865
add a comment |
add a comment |
2
Yes we will prove $a^2+b^2+c^2ge ab+bc+ca$ and it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2ge 0$
– Word Shallow
Dec 19 '18 at 8:22
A more general case including your one can be found here: math.stackexchange.com/questions/1249986/prove-qm-am-inequality
– trancelocation
Dec 19 '18 at 8:25
Another one: Proving that for reals $a,b,c$, $(a + b + c)^2 leq 3(a^2+b^2+c^2)$
– Martin R
Dec 19 '18 at 8:33