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How do I prove the following inequality? [duplicate]
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This question already has an answer here:
Prove that $fracb^2+c^2b+c+fracc^2+a^2c+a+fraca^2+b^2a+b ge a+b+c$
3 answers
How do I proceed to solve the inequality
$$frac(a^2+b^2)(a+b) + frac (b^2+c^2)(b+c) + frac(a^2+c^2)(a+c) geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $frac (a^2+b^2)(a+b)$ to $AM$.
I have tried finding $a_1$ and $a_2$ by doing
$ (frac(a_1+a_2)2bigr)^2 geq$ $frac(a^2+b^2)(a+b)$
algebra-precalculus inequality a.m.-g.m.-inequality
asked Jul 18 at 17:07
mampuuu
304
marked as duplicate by Martin R, José Carlos Santos, abiessu, Ethan Bolker, Parcly Taxel Jul 19 at 1:09
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 |Â
up vote
1
down vote
favorite
This question already has an answer here:
Prove that $fracb^2+c^2b+c+fracc^2+a^2c+a+fraca^2+b^2a+b ge a+b+c$
3 answers
How do I proceed to solve the inequality
$$frac(a^2+b^2)(a+b) + frac (b^2+c^2)(b+c) + frac(a^2+c^2)(a+c) geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $frac (a^2+b^2)(a+b)$ to $AM$.
I have tried finding $a_1$ and $a_2$ by doing
$ (frac(a_1+a_2)2bigr)^2 geq$ $frac(a^2+b^2)(a+b)$
algebra-precalculus inequality a.m.-g.m.-inequality
asked Jul 18 at 17:07
mampuuu
304
marked as duplicate by Martin R, José Carlos Santos, abiessu, Ethan Bolker, Parcly Taxel Jul 19 at 1:09
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
Are the $a,b,c$ positive reals?
– Dr. Sonnhard Graubner
Jul 18 at 17:20
Yes , I am really sorry I forgot to mention that.
– mampuuu
Jul 18 at 17:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Prove that $fracb^2+c^2b+c+fracc^2+a^2c+a+fraca^2+b^2a+b ge a+b+c$
3 answers
How do I proceed to solve the inequality
$$frac(a^2+b^2)(a+b) + frac (b^2+c^2)(b+c) + frac(a^2+c^2)(a+c) geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $frac (a^2+b^2)(a+b)$ to $AM$.
I have tried finding $a_1$ and $a_2$ by doing
$ (frac(a_1+a_2)2bigr)^2 geq$ $frac(a^2+b^2)(a+b)$
algebra-precalculus inequality a.m.-g.m.-inequality
asked Jul 18 at 17:07
mampuuu
304
This question already has an answer here:
Prove that $fracb^2+c^2b+c+fracc^2+a^2c+a+fraca^2+b^2a+b ge a+b+c$
3 answers
How do I proceed to solve the inequality
$$frac(a^2+b^2)(a+b) + frac (b^2+c^2)(b+c) + frac(a^2+c^2)(a+c) geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $frac (a^2+b^2)(a+b)$ to $AM$.
I have tried finding $a_1$ and $a_2$ by doing
$ (frac(a_1+a_2)2bigr)^2 geq$ $frac(a^2+b^2)(a+b)$
This question already has an answer here:
Prove that $fracb^2+c^2b+c+fracc^2+a^2c+a+fraca^2+b^2a+b ge a+b+c$
3 answers
algebra-precalculus inequality a.m.-g.m.-inequality
asked Jul 18 at 17:07
mampuuu
304
edited Jul 18 at 17:33
asked Jul 18 at 17:07
mampuuu
304
asked Jul 18 at 17:07
mampuuu
304
asked Jul 18 at 17:07
mampuuu
304
304
marked as duplicate by Martin R, José Carlos Santos, abiessu, Ethan Bolker, Parcly Taxel Jul 19 at 1:09
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, José Carlos Santos, abiessu, Ethan Bolker, Parcly Taxel Jul 19 at 1:09
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
Are the $a,b,c$ positive reals?
– Dr. Sonnhard Graubner
Jul 18 at 17:20
Yes , I am really sorry I forgot to mention that.
– mampuuu
Jul 18 at 17:32
add a comment |Â
2
Are the $a,b,c$ positive reals?
– Dr. Sonnhard Graubner
Jul 18 at 17:20
Yes , I am really sorry I forgot to mention that.
– mampuuu
Jul 18 at 17:32
2
2
Are the $a,b,c$ positive reals?
– Dr. Sonnhard Graubner
Jul 18 at 17:20
Are the $a,b,c$ positive reals?
– Dr. Sonnhard Graubner
Jul 18 at 17:20
Yes , I am really sorry I forgot to mention that.
– mampuuu
Jul 18 at 17:32
Yes , I am really sorry I forgot to mention that.
– mampuuu
Jul 18 at 17:32
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
If $a$, $b$, $c>0$ then
$$fraca^2+b^2a+bgefraca+b2$$
etc.
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
1
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
add a comment |Â
up vote
0
down vote
Using Cauchy Schwarz in Engel form, we have
$$!!fraca^2a+b+fracb^2b+c+fracc^2a+c+
fracb^2a+b+fracc^2b+c+fraca^2a+cgeq frac(a+b+c)^22(a+b+c)+frac(b+c+a)^22(a+b+c)= a+b+c$$
edited Jul 18 at 18:14
Math Lover
12.4k21232
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $a$, $b$, $c>0$ then
$$fraca^2+b^2a+bgefraca+b2$$
etc.
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
1
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
add a comment |Â
up vote
1
down vote
accepted
If $a$, $b$, $c>0$ then
$$fraca^2+b^2a+bgefraca+b2$$
etc.
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
1
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $a$, $b$, $c>0$ then
$$fraca^2+b^2a+bgefraca+b2$$
etc.
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
If $a$, $b$, $c>0$ then
$$fraca^2+b^2a+bgefraca+b2$$
etc.
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
answered Jul 18 at 17:23
Lord Shark the Unknown
85.5k951112
85.5k951112
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
1
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
add a comment |Â
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
1
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
Sir , how did you get that identity?Is that $AM-GM$ inequality?
– mampuuu
Jul 18 at 17:35
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
I don't think so, but it's related to Cauchy-Schwarz. It's quite easy to prove from scratch.
– Lord Shark the Unknown
Jul 18 at 17:37
1
1
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
@mampuuu This inequality is the quadratic-arithmetic mean inequality in disguise.
– Math Lover
Jul 18 at 17:41
add a comment |Â
up vote
0
down vote
Using Cauchy Schwarz in Engel form, we have
$$!!fraca^2a+b+fracb^2b+c+fracc^2a+c+
fracb^2a+b+fracc^2b+c+fraca^2a+cgeq frac(a+b+c)^22(a+b+c)+frac(b+c+a)^22(a+b+c)= a+b+c$$
edited Jul 18 at 18:14
Math Lover
12.4k21232
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
add a comment |Â
up vote
0
down vote
Using Cauchy Schwarz in Engel form, we have
$$!!fraca^2a+b+fracb^2b+c+fracc^2a+c+
fracb^2a+b+fracc^2b+c+fraca^2a+cgeq frac(a+b+c)^22(a+b+c)+frac(b+c+a)^22(a+b+c)= a+b+c$$
edited Jul 18 at 18:14
Math Lover
12.4k21232
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Using Cauchy Schwarz in Engel form, we have
$$!!fraca^2a+b+fracb^2b+c+fracc^2a+c+
fracb^2a+b+fracc^2b+c+fraca^2a+cgeq frac(a+b+c)^22(a+b+c)+frac(b+c+a)^22(a+b+c)= a+b+c$$
edited Jul 18 at 18:14
Math Lover
12.4k21232
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
Using Cauchy Schwarz in Engel form, we have
$$!!fraca^2a+b+fracb^2b+c+fracc^2a+c+
fracb^2a+b+fracc^2b+c+fraca^2a+cgeq frac(a+b+c)^22(a+b+c)+frac(b+c+a)^22(a+b+c)= a+b+c$$
edited Jul 18 at 18:14
Math Lover
12.4k21232
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
edited Jul 18 at 18:14
Math Lover
12.4k21232
edited Jul 18 at 18:14
Math Lover
12.4k21232
edited Jul 18 at 18:14
Math Lover
12.4k21232
12.4k21232
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
answered Jul 18 at 17:27
Dr. Sonnhard Graubner
66.8k32659
66.8k32659
add a comment |Â
add a comment |Â
2
Are the $a,b,c$ positive reals?
– Dr. Sonnhard Graubner
Jul 18 at 17:20
Yes , I am really sorry I forgot to mention that.
– mampuuu
Jul 18 at 17:32