Can we prove this inequality holds for some region?












1












$begingroup$


Can we prove this inequality holds form some region
, if $(x-a)^2+(y-b)^2>R^2$ then
$$
begin{align}
-xarctan(x)+0.5x-yarctan(y)+0.4y<0\
end{align}
$$
As I can see from the WolframAlpha picture there is some region that is similar to $(x-a)^2+(y-b)^2>R^2$ but I have no idea how we can prove that and find $R$



enter image description here






inequality







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Can we prove this inequality holds form some region
    , if $(x-a)^2+(y-b)^2>R^2$ then
    $$
    begin{align}
    -xarctan(x)+0.5x-yarctan(y)+0.4y<0\
    end{align}
    $$
    As I can see from the WolframAlpha picture there is some region that is similar to $(x-a)^2+(y-b)^2>R^2$ but I have no idea how we can prove that and find $R$



    enter image description here






    inequality







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Can we prove this inequality holds form some region
      , if $(x-a)^2+(y-b)^2>R^2$ then
      $$
      begin{align}
      -xarctan(x)+0.5x-yarctan(y)+0.4y<0\
      end{align}
      $$
      As I can see from the WolframAlpha picture there is some region that is similar to $(x-a)^2+(y-b)^2>R^2$ but I have no idea how we can prove that and find $R$



      enter image description here






      inequality







      share|cite|improve this question









      $endgroup$




      Can we prove this inequality holds form some region
      , if $(x-a)^2+(y-b)^2>R^2$ then
      $$
      begin{align}
      -xarctan(x)+0.5x-yarctan(y)+0.4y<0\
      end{align}
      $$
      As I can see from the WolframAlpha picture there is some region that is similar to $(x-a)^2+(y-b)^2>R^2$ but I have no idea how we can prove that and find $R$



      enter image description here






      inequality




      inequality






      share|cite|improve this question













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      asked Dec 2 '18 at 18:04









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          1 Answer
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          0












          $begingroup$

          The picture suggests to take $a=0.5/2$, $b=0.4/2$ and $Rsimeq 1/3$. We can bound $R$ more precisely
          looking for a big disk $D$ with this center $(a,b)$, which is contained in the white region. Thus we have to show that if $(x,y)in D$ then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge 0.$$



          We use a fact that for each non-negative $t$ we have $operatorname{arctan} tle t$. Therefore we have $-toperatorname{arctan} tge –t^2$ for each real $t$. Then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge$$ $$–x^2+0.5x-y^2+0.4y=$$ $$-(x-a)^2+a^2-(y-b)^2+b^2ge a^2+b^2-R^2.$$



          So $R=sqrt{a^2+b^2}=sqrt{0.1025}=0.3201dots.$ fits.



          I thank to user8053696 for providing the picture:



          enter image description here



          But to find a small disk which contains the white region, we have to look for other, I think, not so simple arguments.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
            $endgroup$
            – Tag
            Dec 3 '18 at 7:51












          • $begingroup$
            For example, from 5>-5 doesn't follow that 5 is negative.
            $endgroup$
            – Tag
            Dec 3 '18 at 7:57










          • $begingroup$
            @user8053696 Thanks for the picture, I clarified my answer.
            $endgroup$
            – Alex Ravsky
            Dec 3 '18 at 13:31











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          1 Answer
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          1 Answer
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          0












          $begingroup$

          The picture suggests to take $a=0.5/2$, $b=0.4/2$ and $Rsimeq 1/3$. We can bound $R$ more precisely
          looking for a big disk $D$ with this center $(a,b)$, which is contained in the white region. Thus we have to show that if $(x,y)in D$ then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge 0.$$



          We use a fact that for each non-negative $t$ we have $operatorname{arctan} tle t$. Therefore we have $-toperatorname{arctan} tge –t^2$ for each real $t$. Then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge$$ $$–x^2+0.5x-y^2+0.4y=$$ $$-(x-a)^2+a^2-(y-b)^2+b^2ge a^2+b^2-R^2.$$



          So $R=sqrt{a^2+b^2}=sqrt{0.1025}=0.3201dots.$ fits.



          I thank to user8053696 for providing the picture:



          enter image description here



          But to find a small disk which contains the white region, we have to look for other, I think, not so simple arguments.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
            $endgroup$
            – Tag
            Dec 3 '18 at 7:51












          • $begingroup$
            For example, from 5>-5 doesn't follow that 5 is negative.
            $endgroup$
            – Tag
            Dec 3 '18 at 7:57










          • $begingroup$
            @user8053696 Thanks for the picture, I clarified my answer.
            $endgroup$
            – Alex Ravsky
            Dec 3 '18 at 13:31
















          0












          $begingroup$

          The picture suggests to take $a=0.5/2$, $b=0.4/2$ and $Rsimeq 1/3$. We can bound $R$ more precisely
          looking for a big disk $D$ with this center $(a,b)$, which is contained in the white region. Thus we have to show that if $(x,y)in D$ then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge 0.$$



          We use a fact that for each non-negative $t$ we have $operatorname{arctan} tle t$. Therefore we have $-toperatorname{arctan} tge –t^2$ for each real $t$. Then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge$$ $$–x^2+0.5x-y^2+0.4y=$$ $$-(x-a)^2+a^2-(y-b)^2+b^2ge a^2+b^2-R^2.$$



          So $R=sqrt{a^2+b^2}=sqrt{0.1025}=0.3201dots.$ fits.



          I thank to user8053696 for providing the picture:



          enter image description here



          But to find a small disk which contains the white region, we have to look for other, I think, not so simple arguments.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
            $endgroup$
            – Tag
            Dec 3 '18 at 7:51












          • $begingroup$
            For example, from 5>-5 doesn't follow that 5 is negative.
            $endgroup$
            – Tag
            Dec 3 '18 at 7:57










          • $begingroup$
            @user8053696 Thanks for the picture, I clarified my answer.
            $endgroup$
            – Alex Ravsky
            Dec 3 '18 at 13:31














          0












          0








          0





          $begingroup$

          The picture suggests to take $a=0.5/2$, $b=0.4/2$ and $Rsimeq 1/3$. We can bound $R$ more precisely
          looking for a big disk $D$ with this center $(a,b)$, which is contained in the white region. Thus we have to show that if $(x,y)in D$ then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge 0.$$



          We use a fact that for each non-negative $t$ we have $operatorname{arctan} tle t$. Therefore we have $-toperatorname{arctan} tge –t^2$ for each real $t$. Then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge$$ $$–x^2+0.5x-y^2+0.4y=$$ $$-(x-a)^2+a^2-(y-b)^2+b^2ge a^2+b^2-R^2.$$



          So $R=sqrt{a^2+b^2}=sqrt{0.1025}=0.3201dots.$ fits.



          I thank to user8053696 for providing the picture:



          enter image description here



          But to find a small disk which contains the white region, we have to look for other, I think, not so simple arguments.






          share|cite|improve this answer











          $endgroup$



          The picture suggests to take $a=0.5/2$, $b=0.4/2$ and $Rsimeq 1/3$. We can bound $R$ more precisely
          looking for a big disk $D$ with this center $(a,b)$, which is contained in the white region. Thus we have to show that if $(x,y)in D$ then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge 0.$$



          We use a fact that for each non-negative $t$ we have $operatorname{arctan} tle t$. Therefore we have $-toperatorname{arctan} tge –t^2$ for each real $t$. Then



          $$-xarctan(x)+0.5x-yarctan(y)+0.4yge$$ $$–x^2+0.5x-y^2+0.4y=$$ $$-(x-a)^2+a^2-(y-b)^2+b^2ge a^2+b^2-R^2.$$



          So $R=sqrt{a^2+b^2}=sqrt{0.1025}=0.3201dots.$ fits.



          I thank to user8053696 for providing the picture:



          enter image description here



          But to find a small disk which contains the white region, we have to look for other, I think, not so simple arguments.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 13:31

























          answered Dec 3 '18 at 6:23









          Alex RavskyAlex Ravsky

          39.6k32181




          39.6k32181












          • $begingroup$
            Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
            $endgroup$
            – Tag
            Dec 3 '18 at 7:51












          • $begingroup$
            For example, from 5>-5 doesn't follow that 5 is negative.
            $endgroup$
            – Tag
            Dec 3 '18 at 7:57










          • $begingroup$
            @user8053696 Thanks for the picture, I clarified my answer.
            $endgroup$
            – Alex Ravsky
            Dec 3 '18 at 13:31


















          • $begingroup$
            Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
            $endgroup$
            – Tag
            Dec 3 '18 at 7:51












          • $begingroup$
            For example, from 5>-5 doesn't follow that 5 is negative.
            $endgroup$
            – Tag
            Dec 3 '18 at 7:57










          • $begingroup$
            @user8053696 Thanks for the picture, I clarified my answer.
            $endgroup$
            – Alex Ravsky
            Dec 3 '18 at 13:31
















          $begingroup$
          Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
          $endgroup$
          – Tag
          Dec 3 '18 at 7:51






          $begingroup$
          Could you, please, clarify why if $-x^2+0.5x-y^2+0.4y<0$ then $-xarctan(x)+0.5x-yarctan(y)+0.4y<0$? Because this circle will be inside closed region of function with arctan not outside. i.imgur.com/n9Xy2z1.png
          $endgroup$
          – Tag
          Dec 3 '18 at 7:51














          $begingroup$
          For example, from 5>-5 doesn't follow that 5 is negative.
          $endgroup$
          – Tag
          Dec 3 '18 at 7:57




          $begingroup$
          For example, from 5>-5 doesn't follow that 5 is negative.
          $endgroup$
          – Tag
          Dec 3 '18 at 7:57












          $begingroup$
          @user8053696 Thanks for the picture, I clarified my answer.
          $endgroup$
          – Alex Ravsky
          Dec 3 '18 at 13:31




          $begingroup$
          @user8053696 Thanks for the picture, I clarified my answer.
          $endgroup$
          – Alex Ravsky
          Dec 3 '18 at 13:31


















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