Little inequality
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After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not always true like this:
$4(x^2z + y^2x + z^2y)≥12xyz$ from MA - MG
$x^3 + y^3 + z^3≥x^2y + y^2z+z^2x$
and this remains:
$x^2z + y^2x + z^2y≥x^2y + y^2z+z^2x$ which doesn't always hold? or does it?
inequality buffalo-way
$endgroup$
add a comment |
$begingroup$
After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not always true like this:
$4(x^2z + y^2x + z^2y)≥12xyz$ from MA - MG
$x^3 + y^3 + z^3≥x^2y + y^2z+z^2x$
and this remains:
$x^2z + y^2x + z^2y≥x^2y + y^2z+z^2x$ which doesn't always hold? or does it?
inequality buffalo-way
$endgroup$
$begingroup$
You're assuming $x,y,z>0$? Otherwise it's clearly wrong, as I can just send $x$ off to $-infty$.
$endgroup$
– Richard Martin
Nov 30 '18 at 11:55
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Yes , I forgot to say that
$endgroup$
– reducere cs
Nov 30 '18 at 11:56
$begingroup$
Wait I think I wrote it wrong
$endgroup$
– reducere cs
Nov 30 '18 at 12:01
$begingroup$
Now it's good check.
$endgroup$
– reducere cs
Nov 30 '18 at 12:02
add a comment |
$begingroup$
After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not always true like this:
$4(x^2z + y^2x + z^2y)≥12xyz$ from MA - MG
$x^3 + y^3 + z^3≥x^2y + y^2z+z^2x$
and this remains:
$x^2z + y^2x + z^2y≥x^2y + y^2z+z^2x$ which doesn't always hold? or does it?
inequality buffalo-way
$endgroup$
After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not always true like this:
$4(x^2z + y^2x + z^2y)≥12xyz$ from MA - MG
$x^3 + y^3 + z^3≥x^2y + y^2z+z^2x$
and this remains:
$x^2z + y^2x + z^2y≥x^2y + y^2z+z^2x$ which doesn't always hold? or does it?
inequality buffalo-way
inequality buffalo-way
edited Nov 30 '18 at 13:10
Michael Rozenberg
98.5k1590189
98.5k1590189
asked Nov 30 '18 at 11:48
reducere csreducere cs
11
11
$begingroup$
You're assuming $x,y,z>0$? Otherwise it's clearly wrong, as I can just send $x$ off to $-infty$.
$endgroup$
– Richard Martin
Nov 30 '18 at 11:55
$begingroup$
Yes , I forgot to say that
$endgroup$
– reducere cs
Nov 30 '18 at 11:56
$begingroup$
Wait I think I wrote it wrong
$endgroup$
– reducere cs
Nov 30 '18 at 12:01
$begingroup$
Now it's good check.
$endgroup$
– reducere cs
Nov 30 '18 at 12:02
add a comment |
$begingroup$
You're assuming $x,y,z>0$? Otherwise it's clearly wrong, as I can just send $x$ off to $-infty$.
$endgroup$
– Richard Martin
Nov 30 '18 at 11:55
$begingroup$
Yes , I forgot to say that
$endgroup$
– reducere cs
Nov 30 '18 at 11:56
$begingroup$
Wait I think I wrote it wrong
$endgroup$
– reducere cs
Nov 30 '18 at 12:01
$begingroup$
Now it's good check.
$endgroup$
– reducere cs
Nov 30 '18 at 12:02
$begingroup$
You're assuming $x,y,z>0$? Otherwise it's clearly wrong, as I can just send $x$ off to $-infty$.
$endgroup$
– Richard Martin
Nov 30 '18 at 11:55
$begingroup$
You're assuming $x,y,z>0$? Otherwise it's clearly wrong, as I can just send $x$ off to $-infty$.
$endgroup$
– Richard Martin
Nov 30 '18 at 11:55
$begingroup$
Yes , I forgot to say that
$endgroup$
– reducere cs
Nov 30 '18 at 11:56
$begingroup$
Yes , I forgot to say that
$endgroup$
– reducere cs
Nov 30 '18 at 11:56
$begingroup$
Wait I think I wrote it wrong
$endgroup$
– reducere cs
Nov 30 '18 at 12:01
$begingroup$
Wait I think I wrote it wrong
$endgroup$
– reducere cs
Nov 30 '18 at 12:01
$begingroup$
Now it's good check.
$endgroup$
– reducere cs
Nov 30 '18 at 12:02
$begingroup$
Now it's good check.
$endgroup$
– reducere cs
Nov 30 '18 at 12:02
add a comment |
1 Answer
1
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votes
$begingroup$
Buffalo Way helps here very well.
Indeed, let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$sum_{cyc}4x^2zgeq12xyz$$ it's enough to prove that
$$sum_{cyc}(x^3+2x^2z-3x^2y)geq0,$$ which we can prove by BW again:
Let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3geq0$$
and we are done!
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Buffalo Way helps here very well.
Indeed, let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$sum_{cyc}4x^2zgeq12xyz$$ it's enough to prove that
$$sum_{cyc}(x^3+2x^2z-3x^2y)geq0,$$ which we can prove by BW again:
Let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3geq0$$
and we are done!
$endgroup$
add a comment |
$begingroup$
Buffalo Way helps here very well.
Indeed, let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$sum_{cyc}4x^2zgeq12xyz$$ it's enough to prove that
$$sum_{cyc}(x^3+2x^2z-3x^2y)geq0,$$ which we can prove by BW again:
Let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3geq0$$
and we are done!
$endgroup$
add a comment |
$begingroup$
Buffalo Way helps here very well.
Indeed, let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$sum_{cyc}4x^2zgeq12xyz$$ it's enough to prove that
$$sum_{cyc}(x^3+2x^2z-3x^2y)geq0,$$ which we can prove by BW again:
Let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3geq0$$
and we are done!
$endgroup$
Buffalo Way helps here very well.
Indeed, let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$sum_{cyc}4x^2zgeq12xyz$$ it's enough to prove that
$$sum_{cyc}(x^3+2x^2z-3x^2y)geq0,$$ which we can prove by BW again:
Let $x=min{x,y,z}$, $y=x+u$ and $z=x+v$.
Thus,
$$sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3geq0$$
and we are done!
edited Nov 30 '18 at 12:55
answered Nov 30 '18 at 12:33
Michael RozenbergMichael Rozenberg
98.5k1590189
98.5k1590189
add a comment |
add a comment |
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$begingroup$
You're assuming $x,y,z>0$? Otherwise it's clearly wrong, as I can just send $x$ off to $-infty$.
$endgroup$
– Richard Martin
Nov 30 '18 at 11:55
$begingroup$
Yes , I forgot to say that
$endgroup$
– reducere cs
Nov 30 '18 at 11:56
$begingroup$
Wait I think I wrote it wrong
$endgroup$
– reducere cs
Nov 30 '18 at 12:01
$begingroup$
Now it's good check.
$endgroup$
– reducere cs
Nov 30 '18 at 12:02