$4x^{4} + 4y^{3} + 5x^{2} + y + 1geq 12xy$ for all positive $x,y$
$begingroup$
Prove this inequality: $$4x^4 + 4y^3 + 5x^2 + y + 1 geq 12xy$$ if $x$ and $y$ are real and positive.
Please, I am a beginner and have no idea how to solve this, so don't use any strange theorems.
algebra-precalculus inequality
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
asked Dec 8 '14 at 18:18
VriskVrisk
12012
$endgroup$
add a comment |
$begingroup$
Prove this inequality: $$4x^4 + 4y^3 + 5x^2 + y + 1 geq 12xy$$ if $x$ and $y$ are real and positive.
Please, I am a beginner and have no idea how to solve this, so don't use any strange theorems.
algebra-precalculus inequality
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
asked Dec 8 '14 at 18:18
VriskVrisk
12012
$endgroup$
$begingroup$
from a graph it looks like it might be enough that $y>0$
$endgroup$
– Mirko
Dec 8 '14 at 18:31
$begingroup$
Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$.
$endgroup$
– Milly
Dec 8 '14 at 18:38
add a comment |
$begingroup$
Prove this inequality: $$4x^4 + 4y^3 + 5x^2 + y + 1 geq 12xy$$ if $x$ and $y$ are real and positive.
Please, I am a beginner and have no idea how to solve this, so don't use any strange theorems.
algebra-precalculus inequality
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
asked Dec 8 '14 at 18:18
VriskVrisk
12012
$endgroup$
Prove this inequality: $$4x^4 + 4y^3 + 5x^2 + y + 1 geq 12xy$$ if $x$ and $y$ are real and positive.
Please, I am a beginner and have no idea how to solve this, so don't use any strange theorems.
algebra-precalculus inequality
algebra-precalculus inequality
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
asked Dec 8 '14 at 18:18
VriskVrisk
12012
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
asked Dec 8 '14 at 18:18
VriskVrisk
12012
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
edited Jan 28 '15 at 2:16
Alex Ravsky
43.5k32583
43.5k32583
asked Dec 8 '14 at 18:18
VriskVrisk
12012
asked Dec 8 '14 at 18:18
VriskVrisk
12012
asked Dec 8 '14 at 18:18
VriskVrisk
12012
12012
$begingroup$
from a graph it looks like it might be enough that $y>0$
$endgroup$
– Mirko
Dec 8 '14 at 18:31
$begingroup$
Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$.
$endgroup$
– Milly
Dec 8 '14 at 18:38
add a comment |
$begingroup$
from a graph it looks like it might be enough that $y>0$
$endgroup$
– Mirko
Dec 8 '14 at 18:31
$begingroup$
Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$.
$endgroup$
– Milly
Dec 8 '14 at 18:38
$begingroup$
from a graph it looks like it might be enough that $y>0$
$endgroup$
– Mirko
Dec 8 '14 at 18:31
$begingroup$
from a graph it looks like it might be enough that $y>0$
$endgroup$
– Mirko
Dec 8 '14 at 18:31
$begingroup$
Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$.
$endgroup$
– Milly
Dec 8 '14 at 18:38
$begingroup$
Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$.
$endgroup$
– Milly
Dec 8 '14 at 18:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I shall use only the inequality $a+bge 2sqrt{ab}.$ We have
$$4x^4+1ge 2sqrt{4x^4cdot 1}=4x^2,$$
$$4y^3+yge 2sqrt{4y^3cdot y}=4y^2.$$
Then
$$4x^4 + 4y^3 + 5x^2 + y + 1ge 4x^2+4y^2+5x^2ge 2sqrt{9x^2cdot 4y^2}=12xy.$$
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
$endgroup$
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
I shall use only the inequality $a+bge 2sqrt{ab}.$ We have
$$4x^4+1ge 2sqrt{4x^4cdot 1}=4x^2,$$
$$4y^3+yge 2sqrt{4y^3cdot y}=4y^2.$$
Then
$$4x^4 + 4y^3 + 5x^2 + y + 1ge 4x^2+4y^2+5x^2ge 2sqrt{9x^2cdot 4y^2}=12xy.$$
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
$endgroup$
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
add a comment |
$begingroup$
I shall use only the inequality $a+bge 2sqrt{ab}.$ We have
$$4x^4+1ge 2sqrt{4x^4cdot 1}=4x^2,$$
$$4y^3+yge 2sqrt{4y^3cdot y}=4y^2.$$
Then
$$4x^4 + 4y^3 + 5x^2 + y + 1ge 4x^2+4y^2+5x^2ge 2sqrt{9x^2cdot 4y^2}=12xy.$$
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
$endgroup$
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
add a comment |
$begingroup$
I shall use only the inequality $a+bge 2sqrt{ab}.$ We have
$$4x^4+1ge 2sqrt{4x^4cdot 1}=4x^2,$$
$$4y^3+yge 2sqrt{4y^3cdot y}=4y^2.$$
Then
$$4x^4 + 4y^3 + 5x^2 + y + 1ge 4x^2+4y^2+5x^2ge 2sqrt{9x^2cdot 4y^2}=12xy.$$
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
$endgroup$
I shall use only the inequality $a+bge 2sqrt{ab}.$ We have
$$4x^4+1ge 2sqrt{4x^4cdot 1}=4x^2,$$
$$4y^3+yge 2sqrt{4y^3cdot y}=4y^2.$$
Then
$$4x^4 + 4y^3 + 5x^2 + y + 1ge 4x^2+4y^2+5x^2ge 2sqrt{9x^2cdot 4y^2}=12xy.$$
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
answered Jan 28 '15 at 2:06
Alex RavskyAlex Ravsky
43.5k32583
43.5k32583
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
add a comment |
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
$begingroup$
AM-GM for the win!
$endgroup$
– Daniel W. Farlow
Jan 28 '15 at 2:10
add a comment |
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$begingroup$
from a graph it looks like it might be enough that $y>0$
$endgroup$
– Mirko
Dec 8 '14 at 18:31
$begingroup$
Sure, if it is true for $y>0$ then $x<0$ would make RHS negative, while LHS is independent of $sgn(x)$.
$endgroup$
– Milly
Dec 8 '14 at 18:38