Mixing
Minimizing in 3 variables
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Find the least possible value of the fraction $dfraca^2+b^2+c^2ab+bc$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3sqrt[3]-2ac/b(a+c)$
And I cant somehow move on.
inequality contest-math
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 8:08
SuperMage1
687210
add a comment |Â
up vote
2
down vote
favorite
Find the least possible value of the fraction $dfraca^2+b^2+c^2ab+bc$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3sqrt[3]-2ac/b(a+c)$
And I cant somehow move on.
inequality contest-math
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 8:08
SuperMage1
687210
You can't apply AM-GM to $2$ positive and $1$ negative variable
– asdf
Jul 16 at 8:10
Do you mean $$fraca^2+b^2+c^2b(a+c)$$?
– Dr. Sonnhard Graubner
Jul 16 at 8:10
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the least possible value of the fraction $dfraca^2+b^2+c^2ab+bc$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3sqrt[3]-2ac/b(a+c)$
And I cant somehow move on.
inequality contest-math
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 8:08
SuperMage1
687210
Find the least possible value of the fraction $dfraca^2+b^2+c^2ab+bc$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3sqrt[3]-2ac/b(a+c)$
And I cant somehow move on.
inequality contest-math
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
asked Jul 16 at 8:08
SuperMage1
687210
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
edited Jul 16 at 8:14
Rodrigo de Azevedo
12.5k41751
12.5k41751
asked Jul 16 at 8:08
SuperMage1
687210
asked Jul 16 at 8:08
SuperMage1
687210
asked Jul 16 at 8:08
SuperMage1
687210
687210
You can't apply AM-GM to $2$ positive and $1$ negative variable
– asdf
Jul 16 at 8:10
Do you mean $$fraca^2+b^2+c^2b(a+c)$$?
– Dr. Sonnhard Graubner
Jul 16 at 8:10
add a comment |Â
You can't apply AM-GM to $2$ positive and $1$ negative variable
– asdf
Jul 16 at 8:10
Do you mean $$fraca^2+b^2+c^2b(a+c)$$?
– Dr. Sonnhard Graubner
Jul 16 at 8:10
You can't apply AM-GM to $2$ positive and $1$ negative variable
– asdf
Jul 16 at 8:10
You can't apply AM-GM to $2$ positive and $1$ negative variable
– asdf
Jul 16 at 8:10
Do you mean $$fraca^2+b^2+c^2b(a+c)$$?
– Dr. Sonnhard Graubner
Jul 16 at 8:10
Do you mean $$fraca^2+b^2+c^2b(a+c)$$?
– Dr. Sonnhard Graubner
Jul 16 at 8:10
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
You can do the following:
$$fraca^2+b^2+c^2b(a+c)geq fracfrac12(a+c)^2+b^2b(a+c)=frac12fraca+cb+fracba+c$$
where at the first step we used $2(a^2+c^2)geq (a+c)^2$
If you now set $fraca+cb=x$ then you just have to minimize $fracx2+frac1x$ over the positive real numbers.
But you have that by AM-GM $fracx2+frac1xgeq sqrt2$ with equality if $x=sqrt2$, i.e. $a+c=sqrt2b$
Going back even further we want $a=c$ from the first step
answered Jul 16 at 8:17
asdf
3,378519
add a comment |Â
up vote
1
down vote
Write
$$frac2a^2+b^2+b^2+2c^22=frac2a^2+b^22+fracb^2+2c^22geq sqrt2ab+sqrt2bc$$
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
add a comment |Â
up vote
1
down vote
Let $dfraca^2+b^2+c^2ab+bc=k>0$ as $a,b,c>0$
$iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)ge0iff k^2gedfrac4(a^2+c^2)(a+c)^2$$
the equality occurs if $a=dfrack(a+c)2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2ge0implies2(a^2+c^2)ge(a+c)^2$
$implies k^2ge2$
the equality occurs if $a=c$
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
add a comment |Â
up vote
1
down vote
Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
fraca^2+b^2+c^2ab+bc = frac 12 left (fraca^2+(2-a)^2b + b
right) ge sqrta^2 + (2-a)^2 = sqrt2(a-1)^2 + 2 , ,
$$
using $AM ge GM$. It follows that
$$
fraca^2+b^2+c^2ab+bc ge sqrt 2 , ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, sqrt 2, 1)$.
answered Jul 16 at 8:59
Martin R
23.9k32743
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You can do the following:
$$fraca^2+b^2+c^2b(a+c)geq fracfrac12(a+c)^2+b^2b(a+c)=frac12fraca+cb+fracba+c$$
where at the first step we used $2(a^2+c^2)geq (a+c)^2$
If you now set $fraca+cb=x$ then you just have to minimize $fracx2+frac1x$ over the positive real numbers.
But you have that by AM-GM $fracx2+frac1xgeq sqrt2$ with equality if $x=sqrt2$, i.e. $a+c=sqrt2b$
Going back even further we want $a=c$ from the first step
answered Jul 16 at 8:17
asdf
3,378519
add a comment |Â
up vote
1
down vote
You can do the following:
$$fraca^2+b^2+c^2b(a+c)geq fracfrac12(a+c)^2+b^2b(a+c)=frac12fraca+cb+fracba+c$$
where at the first step we used $2(a^2+c^2)geq (a+c)^2$
If you now set $fraca+cb=x$ then you just have to minimize $fracx2+frac1x$ over the positive real numbers.
But you have that by AM-GM $fracx2+frac1xgeq sqrt2$ with equality if $x=sqrt2$, i.e. $a+c=sqrt2b$
Going back even further we want $a=c$ from the first step
answered Jul 16 at 8:17
asdf
3,378519
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can do the following:
$$fraca^2+b^2+c^2b(a+c)geq fracfrac12(a+c)^2+b^2b(a+c)=frac12fraca+cb+fracba+c$$
where at the first step we used $2(a^2+c^2)geq (a+c)^2$
If you now set $fraca+cb=x$ then you just have to minimize $fracx2+frac1x$ over the positive real numbers.
But you have that by AM-GM $fracx2+frac1xgeq sqrt2$ with equality if $x=sqrt2$, i.e. $a+c=sqrt2b$
Going back even further we want $a=c$ from the first step
answered Jul 16 at 8:17
asdf
3,378519
You can do the following:
$$fraca^2+b^2+c^2b(a+c)geq fracfrac12(a+c)^2+b^2b(a+c)=frac12fraca+cb+fracba+c$$
where at the first step we used $2(a^2+c^2)geq (a+c)^2$
If you now set $fraca+cb=x$ then you just have to minimize $fracx2+frac1x$ over the positive real numbers.
But you have that by AM-GM $fracx2+frac1xgeq sqrt2$ with equality if $x=sqrt2$, i.e. $a+c=sqrt2b$
Going back even further we want $a=c$ from the first step
answered Jul 16 at 8:17
asdf
3,378519
answered Jul 16 at 8:17
asdf
3,378519
answered Jul 16 at 8:17
asdf
3,378519
answered Jul 16 at 8:17
asdf
3,378519
3,378519
add a comment |Â
add a comment |Â
up vote
1
down vote
Write
$$frac2a^2+b^2+b^2+2c^22=frac2a^2+b^22+fracb^2+2c^22geq sqrt2ab+sqrt2bc$$
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
add a comment |Â
up vote
1
down vote
Write
$$frac2a^2+b^2+b^2+2c^22=frac2a^2+b^22+fracb^2+2c^22geq sqrt2ab+sqrt2bc$$
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Write
$$frac2a^2+b^2+b^2+2c^22=frac2a^2+b^22+fracb^2+2c^22geq sqrt2ab+sqrt2bc$$
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
Write
$$frac2a^2+b^2+b^2+2c^22=frac2a^2+b^22+fracb^2+2c^22geq sqrt2ab+sqrt2bc$$
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
answered Jul 16 at 8:20
Dr. Sonnhard Graubner
66.9k32659
66.9k32659
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
add a comment |Â
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
Elegant and concise solution!
– Martin R
Jul 16 at 9:24
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
I often solved such problems with my students in Leipzig,this is the art of Problem solving see Mathlinks
– Dr. Sonnhard Graubner
Jul 16 at 10:17
add a comment |Â
up vote
1
down vote
Let $dfraca^2+b^2+c^2ab+bc=k>0$ as $a,b,c>0$
$iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)ge0iff k^2gedfrac4(a^2+c^2)(a+c)^2$$
the equality occurs if $a=dfrack(a+c)2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2ge0implies2(a^2+c^2)ge(a+c)^2$
$implies k^2ge2$
the equality occurs if $a=c$
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
add a comment |Â
up vote
1
down vote
Let $dfraca^2+b^2+c^2ab+bc=k>0$ as $a,b,c>0$
$iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)ge0iff k^2gedfrac4(a^2+c^2)(a+c)^2$$
the equality occurs if $a=dfrack(a+c)2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2ge0implies2(a^2+c^2)ge(a+c)^2$
$implies k^2ge2$
the equality occurs if $a=c$
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $dfraca^2+b^2+c^2ab+bc=k>0$ as $a,b,c>0$
$iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)ge0iff k^2gedfrac4(a^2+c^2)(a+c)^2$$
the equality occurs if $a=dfrack(a+c)2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2ge0implies2(a^2+c^2)ge(a+c)^2$
$implies k^2ge2$
the equality occurs if $a=c$
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
Let $dfraca^2+b^2+c^2ab+bc=k>0$ as $a,b,c>0$
$iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)ge0iff k^2gedfrac4(a^2+c^2)(a+c)^2$$
the equality occurs if $a=dfrack(a+c)2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2ge0implies2(a^2+c^2)ge(a+c)^2$
$implies k^2ge2$
the equality occurs if $a=c$
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
edited Jul 16 at 9:10
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
answered Jul 16 at 8:36
lab bhattacharjee
215k14152264
215k14152264
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
add a comment |Â
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
It seems that you mixed up $a$ and $b$ in your calculations, the quadratic equation should be $b^2 - k(a+c)b + a^2 + c^2 = 0$.
– Martin R
Jul 16 at 9:02
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
@MartinR. Thanks for the input. Rectified.
– lab bhattacharjee
Jul 16 at 9:11
add a comment |Â
up vote
1
down vote
Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
fraca^2+b^2+c^2ab+bc = frac 12 left (fraca^2+(2-a)^2b + b
right) ge sqrta^2 + (2-a)^2 = sqrt2(a-1)^2 + 2 , ,
$$
using $AM ge GM$. It follows that
$$
fraca^2+b^2+c^2ab+bc ge sqrt 2 , ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, sqrt 2, 1)$.
answered Jul 16 at 8:59
Martin R
23.9k32743
add a comment |Â
up vote
1
down vote
Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
fraca^2+b^2+c^2ab+bc = frac 12 left (fraca^2+(2-a)^2b + b
right) ge sqrta^2 + (2-a)^2 = sqrt2(a-1)^2 + 2 , ,
$$
using $AM ge GM$. It follows that
$$
fraca^2+b^2+c^2ab+bc ge sqrt 2 , ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, sqrt 2, 1)$.
answered Jul 16 at 8:59
Martin R
23.9k32743
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
fraca^2+b^2+c^2ab+bc = frac 12 left (fraca^2+(2-a)^2b + b
right) ge sqrta^2 + (2-a)^2 = sqrt2(a-1)^2 + 2 , ,
$$
using $AM ge GM$. It follows that
$$
fraca^2+b^2+c^2ab+bc ge sqrt 2 , ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, sqrt 2, 1)$.
answered Jul 16 at 8:59
Martin R
23.9k32743
Yet another approach: The expression does not change if $(a, b, c)$
are multiplied by a common factor, so we can assume that $a+c=2$. Then
$$
fraca^2+b^2+c^2ab+bc = frac 12 left (fraca^2+(2-a)^2b + b
right) ge sqrta^2 + (2-a)^2 = sqrt2(a-1)^2 + 2 , ,
$$
using $AM ge GM$. It follows that
$$
fraca^2+b^2+c^2ab+bc ge sqrt 2 , ,
$$
with equality if and only if $(a, b, c)$ is a multiple of $(1, sqrt 2, 1)$.
answered Jul 16 at 8:59
Martin R
23.9k32743
edited Jul 16 at 9:10
answered Jul 16 at 8:59
Martin R
23.9k32743
answered Jul 16 at 8:59
Martin R
23.9k32743
answered Jul 16 at 8:59
Martin R
23.9k32743
23.9k32743
add a comment |Â
add a comment |Â
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You can't apply AM-GM to $2$ positive and $1$ negative variable
– asdf
Jul 16 at 8:10
Do you mean $$fraca^2+b^2+c^2b(a+c)$$?
– Dr. Sonnhard Graubner
Jul 16 at 8:10