Little inequality












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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










  • $begingroup$
    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














0












0








0


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 13:10









Michael Rozenberg

103k1891195




103k1891195










asked Nov 30 '18 at 11:48









reducere csreducere cs

112




112












  • $begingroup$
    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










  • $begingroup$
    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










1 Answer
1






active

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0












$begingroup$

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!






share|cite|improve this answer











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    $begingroup$

    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!






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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!






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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!






        share|cite|improve this answer











        $endgroup$



        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!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 12:55

























        answered Nov 30 '18 at 12:33









        Michael RozenbergMichael Rozenberg

        103k1891195




        103k1891195






























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