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Minimum value of $fracx^4+5x^2+7x^2+3$
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Minimum value of $$f(x)=fracx^4+5x^2+7x^2+3$$
we have $f(x)$ as
$$f(x)=(x^2+3)+frac1x^2+3-1$$
Now by $AM gt GM$ we have
$$(x^2+3)+frac1x^2+3 gt 2$$
But equality cannot occur since $$x^2+3 ne frac1x^2+3$$
But my question is without using calculus is there any way to find minimum using AM, GM?
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
add a comment |Â
up vote
2
down vote
favorite
Minimum value of $$f(x)=fracx^4+5x^2+7x^2+3$$
we have $f(x)$ as
$$f(x)=(x^2+3)+frac1x^2+3-1$$
Now by $AM gt GM$ we have
$$(x^2+3)+frac1x^2+3 gt 2$$
But equality cannot occur since $$x^2+3 ne frac1x^2+3$$
But my question is without using calculus is there any way to find minimum using AM, GM?
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Minimum value of $$f(x)=fracx^4+5x^2+7x^2+3$$
we have $f(x)$ as
$$f(x)=(x^2+3)+frac1x^2+3-1$$
Now by $AM gt GM$ we have
$$(x^2+3)+frac1x^2+3 gt 2$$
But equality cannot occur since $$x^2+3 ne frac1x^2+3$$
But my question is without using calculus is there any way to find minimum using AM, GM?
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
Minimum value of $$f(x)=fracx^4+5x^2+7x^2+3$$
we have $f(x)$ as
$$f(x)=(x^2+3)+frac1x^2+3-1$$
Now by $AM gt GM$ we have
$$(x^2+3)+frac1x^2+3 gt 2$$
But equality cannot occur since $$x^2+3 ne frac1x^2+3$$
But my question is without using calculus is there any way to find minimum using AM, GM?
algebra-precalculus optimization maxima-minima a.m.-g.m.-inequality
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
edited Jul 31 at 9:58
Michael Rozenberg
87.5k1577179
87.5k1577179
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
asked Jul 31 at 9:25
Ekaveera Kumar Sharma
5,13311122
5,13311122
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
2
down vote
accepted
As an alternative, using your idea for decomposition, by Rearrangement inequality with
- $(a_1,a_2)=(frac13,frac1(x^2+3))$
- $(b_1,b_2)=left(3(x^2+3),1right)$
we have that
$$a_1b_1+a_2b_2=frac13cdot 3(x^2+3)+frac1(x^2+3)cdot 1= x^2+3+frac1(x^2+3)ge a_1b_2+a_2b_1=$$
$$=frac13cdot 1 +frac1(x^2+3)cdot 3(x^2+3)=frac13+3=frac103$$
with equality for
$$a_1=a_2 iff frac13=frac1(x^2+3)iff x=0$$
therefore
$$f(x)=(x^2+3)+frac1x^2+3-1ge frac103-1=frac 73$$
with the minimum attained at $x=0$.
answered Jul 31 at 10:11
gimusi
64.1k73480
add a comment |Â
up vote
2
down vote
Hint: We have
$$fracx^4+5x^2+7x^2+3geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21geq 7x^2+21$$
$$x^2(3x^2+8)geq 0$$ the equal sign holds it $$x=0$$
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
1
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
add a comment |Â
up vote
1
down vote
Let $ageq 1$ be fixed. Then, $g:[a,infty)tomathbbR$ defined by $$g(t):=t+frac1ttext for all tin[a,infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)left(1-frac1atright),.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2,left(fracta^2right)+frac1tgeq left(a^2+1right)left(left(fracta^2right)^a^2,frac1tright)^frac1a^2+1=left(a^2+1right),left(fract^a^2-1a^2a^2right)^frac1a^2+1,,.$$
Since $tgeq ageq1$, we get
$$g(t)geq left(a^2+1right),left(fraca^a^2-1a^2a^2right)^frac1a^2+1=left(a^2+1right),frac1a=g(a),.$$
answered Jul 31 at 9:30
Batominovski
22.8k22776
add a comment |Â
up vote
1
down vote
For $x=0$ we get a value $frac73.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$fracx^4+5x^2+7x^2+3geqfrac73$$ or
$$x^2(3x^2+8)geq0,$$ which is obvious.
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
1
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
 |Â
show 3 more comments
up vote
0
down vote
Let $y=f(x)$, then
beginalign
0 &= x^4+(5-y)x^2+(7-3y) \
x^2 &= fracy-5 pm sqrt(5-y)^2+4(colorred3y-7)2 \
&= fracy-5 pm sqrty^2+2y-32 \
&= fracy-5 pm sqrt(y+3)(y-1)2
endalign
Now,
$$Delta=(y+3)(y-1) ge 0$$
Together with $f(x)>0$,
$$y ge 1$$
in which $Delta$ is increasing with $y$.
- When $y ge 5$,
beginalign
3y-7 & ge 3(5)-7 \
& = 8 \
& > 0 \
Delta & > sqrt(5-y)^2 \
& = y-5 \
endalign
- When $1 le y le 5$,
beginalign
y-5+sqrt(y+3)(y-1) & ge 0 tag$x^2 ge 0$ \
(y+3)(y-1) & ge (5-y)^2 \
4(colorred3y-7) & ge 0 \
y & ge frac73 \
endalign
The non-negativity of $x^2$ imposes $$x^2=fracy-5+sqrt(y+3)(y-1)2$$
which is increasing with $y$.
The required minimum is $frac73$ which is achieved when
$$x^2=fracfrac73-5+5-frac732=0$$
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
As an alternative, using your idea for decomposition, by Rearrangement inequality with
- $(a_1,a_2)=(frac13,frac1(x^2+3))$
- $(b_1,b_2)=left(3(x^2+3),1right)$
we have that
$$a_1b_1+a_2b_2=frac13cdot 3(x^2+3)+frac1(x^2+3)cdot 1= x^2+3+frac1(x^2+3)ge a_1b_2+a_2b_1=$$
$$=frac13cdot 1 +frac1(x^2+3)cdot 3(x^2+3)=frac13+3=frac103$$
with equality for
$$a_1=a_2 iff frac13=frac1(x^2+3)iff x=0$$
therefore
$$f(x)=(x^2+3)+frac1x^2+3-1ge frac103-1=frac 73$$
with the minimum attained at $x=0$.
answered Jul 31 at 10:11
gimusi
64.1k73480
add a comment |Â
up vote
2
down vote
accepted
As an alternative, using your idea for decomposition, by Rearrangement inequality with
- $(a_1,a_2)=(frac13,frac1(x^2+3))$
- $(b_1,b_2)=left(3(x^2+3),1right)$
we have that
$$a_1b_1+a_2b_2=frac13cdot 3(x^2+3)+frac1(x^2+3)cdot 1= x^2+3+frac1(x^2+3)ge a_1b_2+a_2b_1=$$
$$=frac13cdot 1 +frac1(x^2+3)cdot 3(x^2+3)=frac13+3=frac103$$
with equality for
$$a_1=a_2 iff frac13=frac1(x^2+3)iff x=0$$
therefore
$$f(x)=(x^2+3)+frac1x^2+3-1ge frac103-1=frac 73$$
with the minimum attained at $x=0$.
answered Jul 31 at 10:11
gimusi
64.1k73480
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
As an alternative, using your idea for decomposition, by Rearrangement inequality with
- $(a_1,a_2)=(frac13,frac1(x^2+3))$
- $(b_1,b_2)=left(3(x^2+3),1right)$
we have that
$$a_1b_1+a_2b_2=frac13cdot 3(x^2+3)+frac1(x^2+3)cdot 1= x^2+3+frac1(x^2+3)ge a_1b_2+a_2b_1=$$
$$=frac13cdot 1 +frac1(x^2+3)cdot 3(x^2+3)=frac13+3=frac103$$
with equality for
$$a_1=a_2 iff frac13=frac1(x^2+3)iff x=0$$
therefore
$$f(x)=(x^2+3)+frac1x^2+3-1ge frac103-1=frac 73$$
with the minimum attained at $x=0$.
answered Jul 31 at 10:11
gimusi
64.1k73480
As an alternative, using your idea for decomposition, by Rearrangement inequality with
- $(a_1,a_2)=(frac13,frac1(x^2+3))$
- $(b_1,b_2)=left(3(x^2+3),1right)$
we have that
$$a_1b_1+a_2b_2=frac13cdot 3(x^2+3)+frac1(x^2+3)cdot 1= x^2+3+frac1(x^2+3)ge a_1b_2+a_2b_1=$$
$$=frac13cdot 1 +frac1(x^2+3)cdot 3(x^2+3)=frac13+3=frac103$$
with equality for
$$a_1=a_2 iff frac13=frac1(x^2+3)iff x=0$$
therefore
$$f(x)=(x^2+3)+frac1x^2+3-1ge frac103-1=frac 73$$
with the minimum attained at $x=0$.
answered Jul 31 at 10:11
gimusi
64.1k73480
answered Jul 31 at 10:11
gimusi
64.1k73480
answered Jul 31 at 10:11
gimusi
64.1k73480
answered Jul 31 at 10:11
gimusi
64.1k73480
64.1k73480
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: We have
$$fracx^4+5x^2+7x^2+3geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21geq 7x^2+21$$
$$x^2(3x^2+8)geq 0$$ the equal sign holds it $$x=0$$
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
1
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
add a comment |Â
up vote
2
down vote
Hint: We have
$$fracx^4+5x^2+7x^2+3geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21geq 7x^2+21$$
$$x^2(3x^2+8)geq 0$$ the equal sign holds it $$x=0$$
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
1
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: We have
$$fracx^4+5x^2+7x^2+3geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21geq 7x^2+21$$
$$x^2(3x^2+8)geq 0$$ the equal sign holds it $$x=0$$
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
Hint: We have
$$fracx^4+5x^2+7x^2+3geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21geq 7x^2+21$$
$$x^2(3x^2+8)geq 0$$ the equal sign holds it $$x=0$$
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
answered Jul 31 at 9:53
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
1
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
add a comment |Â
1
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
1
1
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:13
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
You have right, we can compute the first derivative additionally
– Dr. Sonnhard Graubner
Jul 31 at 10:15
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
Yes ok but the OP is asking for a method without derivatives. Your way is good but the initial guessing is a weakness point.
– gimusi
Jul 31 at 10:25
add a comment |Â
up vote
1
down vote
Let $ageq 1$ be fixed. Then, $g:[a,infty)tomathbbR$ defined by $$g(t):=t+frac1ttext for all tin[a,infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)left(1-frac1atright),.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2,left(fracta^2right)+frac1tgeq left(a^2+1right)left(left(fracta^2right)^a^2,frac1tright)^frac1a^2+1=left(a^2+1right),left(fract^a^2-1a^2a^2right)^frac1a^2+1,,.$$
Since $tgeq ageq1$, we get
$$g(t)geq left(a^2+1right),left(fraca^a^2-1a^2a^2right)^frac1a^2+1=left(a^2+1right),frac1a=g(a),.$$
answered Jul 31 at 9:30
Batominovski
22.8k22776
add a comment |Â
up vote
1
down vote
Let $ageq 1$ be fixed. Then, $g:[a,infty)tomathbbR$ defined by $$g(t):=t+frac1ttext for all tin[a,infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)left(1-frac1atright),.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2,left(fracta^2right)+frac1tgeq left(a^2+1right)left(left(fracta^2right)^a^2,frac1tright)^frac1a^2+1=left(a^2+1right),left(fract^a^2-1a^2a^2right)^frac1a^2+1,,.$$
Since $tgeq ageq1$, we get
$$g(t)geq left(a^2+1right),left(fraca^a^2-1a^2a^2right)^frac1a^2+1=left(a^2+1right),frac1a=g(a),.$$
answered Jul 31 at 9:30
Batominovski
22.8k22776
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $ageq 1$ be fixed. Then, $g:[a,infty)tomathbbR$ defined by $$g(t):=t+frac1ttext for all tin[a,infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)left(1-frac1atright),.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2,left(fracta^2right)+frac1tgeq left(a^2+1right)left(left(fracta^2right)^a^2,frac1tright)^frac1a^2+1=left(a^2+1right),left(fract^a^2-1a^2a^2right)^frac1a^2+1,,.$$
Since $tgeq ageq1$, we get
$$g(t)geq left(a^2+1right),left(fraca^a^2-1a^2a^2right)^frac1a^2+1=left(a^2+1right),frac1a=g(a),.$$
answered Jul 31 at 9:30
Batominovski
22.8k22776
Let $ageq 1$ be fixed. Then, $g:[a,infty)tomathbbR$ defined by $$g(t):=t+frac1ttext for all tin[a,infty)$$ is minimized at $t=a$. To show this, write
$$g(t)-g(a)=(t-a)left(1-frac1atright),.$$
With the Weighted AM-GM inequality, we note that
$$g(t)=a^2,left(fracta^2right)+frac1tgeq left(a^2+1right)left(left(fracta^2right)^a^2,frac1tright)^frac1a^2+1=left(a^2+1right),left(fract^a^2-1a^2a^2right)^frac1a^2+1,,.$$
Since $tgeq ageq1$, we get
$$g(t)geq left(a^2+1right),left(fraca^a^2-1a^2a^2right)^frac1a^2+1=left(a^2+1right),frac1a=g(a),.$$
answered Jul 31 at 9:30
Batominovski
22.8k22776
edited Jul 31 at 9:39
answered Jul 31 at 9:30
Batominovski
22.8k22776
answered Jul 31 at 9:30
Batominovski
22.8k22776
answered Jul 31 at 9:30
Batominovski
22.8k22776
22.8k22776
add a comment |Â
add a comment |Â
up vote
1
down vote
For $x=0$ we get a value $frac73.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$fracx^4+5x^2+7x^2+3geqfrac73$$ or
$$x^2(3x^2+8)geq0,$$ which is obvious.
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
1
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
 |Â
show 3 more comments
up vote
1
down vote
For $x=0$ we get a value $frac73.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$fracx^4+5x^2+7x^2+3geqfrac73$$ or
$$x^2(3x^2+8)geq0,$$ which is obvious.
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
1
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
For $x=0$ we get a value $frac73.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$fracx^4+5x^2+7x^2+3geqfrac73$$ or
$$x^2(3x^2+8)geq0,$$ which is obvious.
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
For $x=0$ we get a value $frac73.$
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$fracx^4+5x^2+7x^2+3geqfrac73$$ or
$$x^2(3x^2+8)geq0,$$ which is obvious.
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
answered Jul 31 at 9:51
Michael Rozenberg
87.5k1577179
87.5k1577179
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
1
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
 |Â
show 3 more comments
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
1
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
that's nice but we need to guess at first that the minimum is at $x=0$
– gimusi
Jul 31 at 10:12
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes, of course! It's not so hard. After this we need to prove that for $x=0$ we'll get a minimal value, that I made.
– Michael Rozenberg
Jul 31 at 10:18
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
Yes once we guess that the minimum is attained at $x=0$ the method is very good, but we need to guess that and in general it could be not so obvious to obtain. I've tried by rearrangement inequality. You can find of course some method to derive the result without any initial guessing. What about C-S? I would like to see some other method from your side.
– gimusi
Jul 31 at 10:23
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
@gimusi I think here it's a best way.
– Michael Rozenberg
Jul 31 at 10:29
1
1
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
Why someone down voted?
– Michael Rozenberg
Aug 1 at 11:40
 |Â
show 3 more comments
up vote
0
down vote
Let $y=f(x)$, then
beginalign
0 &= x^4+(5-y)x^2+(7-3y) \
x^2 &= fracy-5 pm sqrt(5-y)^2+4(colorred3y-7)2 \
&= fracy-5 pm sqrty^2+2y-32 \
&= fracy-5 pm sqrt(y+3)(y-1)2
endalign
Now,
$$Delta=(y+3)(y-1) ge 0$$
Together with $f(x)>0$,
$$y ge 1$$
in which $Delta$ is increasing with $y$.
- When $y ge 5$,
beginalign
3y-7 & ge 3(5)-7 \
& = 8 \
& > 0 \
Delta & > sqrt(5-y)^2 \
& = y-5 \
endalign
- When $1 le y le 5$,
beginalign
y-5+sqrt(y+3)(y-1) & ge 0 tag$x^2 ge 0$ \
(y+3)(y-1) & ge (5-y)^2 \
4(colorred3y-7) & ge 0 \
y & ge frac73 \
endalign
The non-negativity of $x^2$ imposes $$x^2=fracy-5+sqrt(y+3)(y-1)2$$
which is increasing with $y$.
The required minimum is $frac73$ which is achieved when
$$x^2=fracfrac73-5+5-frac732=0$$
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
add a comment |Â
up vote
0
down vote
Let $y=f(x)$, then
beginalign
0 &= x^4+(5-y)x^2+(7-3y) \
x^2 &= fracy-5 pm sqrt(5-y)^2+4(colorred3y-7)2 \
&= fracy-5 pm sqrty^2+2y-32 \
&= fracy-5 pm sqrt(y+3)(y-1)2
endalign
Now,
$$Delta=(y+3)(y-1) ge 0$$
Together with $f(x)>0$,
$$y ge 1$$
in which $Delta$ is increasing with $y$.
- When $y ge 5$,
beginalign
3y-7 & ge 3(5)-7 \
& = 8 \
& > 0 \
Delta & > sqrt(5-y)^2 \
& = y-5 \
endalign
- When $1 le y le 5$,
beginalign
y-5+sqrt(y+3)(y-1) & ge 0 tag$x^2 ge 0$ \
(y+3)(y-1) & ge (5-y)^2 \
4(colorred3y-7) & ge 0 \
y & ge frac73 \
endalign
The non-negativity of $x^2$ imposes $$x^2=fracy-5+sqrt(y+3)(y-1)2$$
which is increasing with $y$.
The required minimum is $frac73$ which is achieved when
$$x^2=fracfrac73-5+5-frac732=0$$
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $y=f(x)$, then
beginalign
0 &= x^4+(5-y)x^2+(7-3y) \
x^2 &= fracy-5 pm sqrt(5-y)^2+4(colorred3y-7)2 \
&= fracy-5 pm sqrty^2+2y-32 \
&= fracy-5 pm sqrt(y+3)(y-1)2
endalign
Now,
$$Delta=(y+3)(y-1) ge 0$$
Together with $f(x)>0$,
$$y ge 1$$
in which $Delta$ is increasing with $y$.
- When $y ge 5$,
beginalign
3y-7 & ge 3(5)-7 \
& = 8 \
& > 0 \
Delta & > sqrt(5-y)^2 \
& = y-5 \
endalign
- When $1 le y le 5$,
beginalign
y-5+sqrt(y+3)(y-1) & ge 0 tag$x^2 ge 0$ \
(y+3)(y-1) & ge (5-y)^2 \
4(colorred3y-7) & ge 0 \
y & ge frac73 \
endalign
The non-negativity of $x^2$ imposes $$x^2=fracy-5+sqrt(y+3)(y-1)2$$
which is increasing with $y$.
The required minimum is $frac73$ which is achieved when
$$x^2=fracfrac73-5+5-frac732=0$$
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
Let $y=f(x)$, then
beginalign
0 &= x^4+(5-y)x^2+(7-3y) \
x^2 &= fracy-5 pm sqrt(5-y)^2+4(colorred3y-7)2 \
&= fracy-5 pm sqrty^2+2y-32 \
&= fracy-5 pm sqrt(y+3)(y-1)2
endalign
Now,
$$Delta=(y+3)(y-1) ge 0$$
Together with $f(x)>0$,
$$y ge 1$$
in which $Delta$ is increasing with $y$.
- When $y ge 5$,
beginalign
3y-7 & ge 3(5)-7 \
& = 8 \
& > 0 \
Delta & > sqrt(5-y)^2 \
& = y-5 \
endalign
- When $1 le y le 5$,
beginalign
y-5+sqrt(y+3)(y-1) & ge 0 tag$x^2 ge 0$ \
(y+3)(y-1) & ge (5-y)^2 \
4(colorred3y-7) & ge 0 \
y & ge frac73 \
endalign
The non-negativity of $x^2$ imposes $$x^2=fracy-5+sqrt(y+3)(y-1)2$$
which is increasing with $y$.
The required minimum is $frac73$ which is achieved when
$$x^2=fracfrac73-5+5-frac732=0$$
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
answered Jul 31 at 15:38
Ng Chung Tak
12.9k31129
12.9k31129
add a comment |Â
add a comment |Â
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