$x,y,z > 0$. If $xyz geq xy + yz + zx$ prove that $xyz geq 3(x+y+z)$
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up vote
1
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Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.
inequality contest-math symmetric-polynomials sos
add a comment |Â
up vote
1
down vote
favorite
Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.
inequality contest-math symmetric-polynomials sos
From where does it come?
â Dr. Sonnhard Graubner
Aug 27 at 16:40
It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
â Frousse
Aug 27 at 16:45
The "AM-GM inequality" may help.
â John
Aug 27 at 16:47
Hint: Use AM-HM inequality
â Love Invariants
Aug 27 at 16:48
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.
inequality contest-math symmetric-polynomials sos
Everything is in the title. But if possible, It would very nice if you gave indications rather than the solution.
Thanks in advance.
inequality contest-math symmetric-polynomials sos
edited Aug 27 at 17:05
Michael Rozenberg
89.1k1581180
89.1k1581180
asked Aug 27 at 16:39
Frousse
164
164
From where does it come?
â Dr. Sonnhard Graubner
Aug 27 at 16:40
It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
â Frousse
Aug 27 at 16:45
The "AM-GM inequality" may help.
â John
Aug 27 at 16:47
Hint: Use AM-HM inequality
â Love Invariants
Aug 27 at 16:48
add a comment |Â
From where does it come?
â Dr. Sonnhard Graubner
Aug 27 at 16:40
It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
â Frousse
Aug 27 at 16:45
The "AM-GM inequality" may help.
â John
Aug 27 at 16:47
Hint: Use AM-HM inequality
â Love Invariants
Aug 27 at 16:48
From where does it come?
â Dr. Sonnhard Graubner
Aug 27 at 16:40
From where does it come?
â Dr. Sonnhard Graubner
Aug 27 at 16:40
It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
â Frousse
Aug 27 at 16:45
It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
â Frousse
Aug 27 at 16:45
The "AM-GM inequality" may help.
â John
Aug 27 at 16:47
The "AM-GM inequality" may help.
â John
Aug 27 at 16:47
Hint: Use AM-HM inequality
â Love Invariants
Aug 27 at 16:48
Hint: Use AM-HM inequality
â Love Invariants
Aug 27 at 16:48
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
$$dfrac1x+dfrac1y+dfrac1z le 1$$
$$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$
1
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
add a comment |Â
up vote
3
down vote
Let $1/x=a, 1/y=b, 1/z=c$.
Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.
This is simple:
$1geq (a+b+c)^2geq 3(ab+bc+ca)$.
add a comment |Â
up vote
1
down vote
Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
$$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
$$sum_cycz^2(x-y)^2geq0.$$
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$dfrac1x+dfrac1y+dfrac1z le 1$$
$$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$
1
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
add a comment |Â
up vote
3
down vote
accepted
$$dfrac1x+dfrac1y+dfrac1z le 1$$
$$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$
1
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$dfrac1x+dfrac1y+dfrac1z le 1$$
$$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$
$$dfrac1x+dfrac1y+dfrac1z le 1$$
$$3left(dfrac1xy+dfrac1yz+dfrac1zxright)leleft(dfrac1x+dfrac1y+dfrac1zright)^2le 1$$
answered Aug 27 at 16:48
Sergic Primazon
40215
40215
1
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
add a comment |Â
1
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
1
1
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
Well, thank you but that's pretty much the whole solution, don't you think ? ^^
â Frousse
Aug 27 at 16:59
add a comment |Â
up vote
3
down vote
Let $1/x=a, 1/y=b, 1/z=c$.
Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.
This is simple:
$1geq (a+b+c)^2geq 3(ab+bc+ca)$.
add a comment |Â
up vote
3
down vote
Let $1/x=a, 1/y=b, 1/z=c$.
Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.
This is simple:
$1geq (a+b+c)^2geq 3(ab+bc+ca)$.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $1/x=a, 1/y=b, 1/z=c$.
Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.
This is simple:
$1geq (a+b+c)^2geq 3(ab+bc+ca)$.
Let $1/x=a, 1/y=b, 1/z=c$.
Then the problem is: given $a,b,c>0$ such that $a+b+cleq 1$, show that $ab+bc+caleq 1/3$.
This is simple:
$1geq (a+b+c)^2geq 3(ab+bc+ca)$.
answered Aug 27 at 16:50
A. Pongrácz
4,415725
4,415725
add a comment |Â
add a comment |Â
up vote
1
down vote
Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
$$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
$$sum_cycz^2(x-y)^2geq0.$$
add a comment |Â
up vote
1
down vote
Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
$$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
$$sum_cycz^2(x-y)^2geq0.$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
$$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
$$sum_cycz^2(x-y)^2geq0.$$
Since $$1geqfracxy+xz+yzxyz,$$ it's enough to prove that
$$xyzleft(fracxy+xz+yzxyzright)^2geq3(x+y+z)$$ or
$$(xy+xz+yz)^2geq3xyz(x+y+z)$$ or
$$sum_cycz^2(x-y)^2geq0.$$
answered Aug 27 at 17:01
Michael Rozenberg
89.1k1581180
89.1k1581180
add a comment |Â
add a comment |Â
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From where does it come?
â Dr. Sonnhard Graubner
Aug 27 at 16:40
It's written " India 2001" but when I've checked their national Olympiad I couldn't find it ^^.
â Frousse
Aug 27 at 16:45
The "AM-GM inequality" may help.
â John
Aug 27 at 16:47
Hint: Use AM-HM inequality
â Love Invariants
Aug 27 at 16:48