bound on $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert $?












1














Question: How to get a good bound on $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert $?



Context: I want to show that $frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $Bbb R^2 - K$ , the absolute value of the function is small.



Attempt:



$biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert
leq frac{|x+2y-2|}{x^2+y^2+1} leq frac{sqrt{x^2+y^2}+2(sqrt{x^2+y^2})-2}{x^2+y^2+1} leq frac{3sqrt{x^2+y^2}-2}{x^2+y^2}
$



This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?



But I am not sure how to get rid of the $-2$ in the numerator.






real-analysis







share|cite|improve this question





























    1














    Question: How to get a good bound on $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert $?



    Context: I want to show that $frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $Bbb R^2 - K$ , the absolute value of the function is small.



    Attempt:



    $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert
    leq frac{|x+2y-2|}{x^2+y^2+1} leq frac{sqrt{x^2+y^2}+2(sqrt{x^2+y^2})-2}{x^2+y^2+1} leq frac{3sqrt{x^2+y^2}-2}{x^2+y^2}
    $



    This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?



    But I am not sure how to get rid of the $-2$ in the numerator.






    real-analysis







    share|cite|improve this question



























      1












      1








      1


      2





      Question: How to get a good bound on $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert $?



      Context: I want to show that $frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $Bbb R^2 - K$ , the absolute value of the function is small.



      Attempt:



      $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert
      leq frac{|x+2y-2|}{x^2+y^2+1} leq frac{sqrt{x^2+y^2}+2(sqrt{x^2+y^2})-2}{x^2+y^2+1} leq frac{3sqrt{x^2+y^2}-2}{x^2+y^2}
      $



      This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?



      But I am not sure how to get rid of the $-2$ in the numerator.






      real-analysis







      share|cite|improve this question















      Question: How to get a good bound on $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert $?



      Context: I want to show that $frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $Bbb R^2 - K$ , the absolute value of the function is small.



      Attempt:



      $biggrrvert frac{x+2y-2}{x^2+y^2+1} biggrlvert
      leq frac{|x+2y-2|}{x^2+y^2+1} leq frac{sqrt{x^2+y^2}+2(sqrt{x^2+y^2})-2}{x^2+y^2+1} leq frac{3sqrt{x^2+y^2}-2}{x^2+y^2}
      $



      This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?



      But I am not sure how to get rid of the $-2$ in the numerator.






      real-analysis




      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 '18 at 3:24









      Tianlalu

      3,09621038




      3,09621038










      asked Nov 20 '18 at 2:45









      eatfood

      1827




      1827






















          2 Answers
          2






          active

          oldest

          votes


















          3














          For the upper bound observe
          begin{align}
          c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
          =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} geq 0
          end{align}
          for all $x, y$ iff
          begin{align}
          2+frac{4c^2-5}{4c}=0 Leftrightarrow c=frac{1}{2}.
          end{align}
          Hence it follows
          begin{align}
          x+2y-2 leq frac{1}{2}(x^2+y^2+1) Longleftrightarrow frac{x+2y-2}{x^2+y^2+1}leq frac{1}{2}
          end{align}
          for all $x, y$. Maximum is attained when $(x, y) = (1, 2)$.



          For the lower bound, we consider
          begin{align}
          c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
          =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} leq 0
          end{align}
          which needs to hold for all $x, y$. In particular, the inequality has to hold when $x=frac{1}{2c}$ and $y=frac{1}{c}$. Hence
          begin{align}
          2+frac{4c^2-5}{4c}leq 0
          end{align}
          if $c in (-infty, -5/2]$. Hence
          begin{align}
          -frac{5}{2} leq frac{x+2y-2}{x^2+y^2+1}.
          end{align}






          share|cite|improve this answer





















          • +1 Wow nice technique
            – Ovi
            Nov 20 '18 at 5:09



















          0














          How about



          $$left| dfrac {x+2y-2}{x^2+y^2+1} right| le dfrac {|x|+|2y|+|2|}{|x|+|y|+1} le dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$



          This should hold when $|x|, |y| > 1$.



          You can take $K$ to be the unit square. You can check that $fleft(- dfrac 14, - dfrac 12 right) < -2$, so indeed the function has an absolute minimum.






          share|cite|improve this answer























            Your Answer





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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            For the upper bound observe
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} geq 0
            end{align}
            for all $x, y$ iff
            begin{align}
            2+frac{4c^2-5}{4c}=0 Leftrightarrow c=frac{1}{2}.
            end{align}
            Hence it follows
            begin{align}
            x+2y-2 leq frac{1}{2}(x^2+y^2+1) Longleftrightarrow frac{x+2y-2}{x^2+y^2+1}leq frac{1}{2}
            end{align}
            for all $x, y$. Maximum is attained when $(x, y) = (1, 2)$.



            For the lower bound, we consider
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} leq 0
            end{align}
            which needs to hold for all $x, y$. In particular, the inequality has to hold when $x=frac{1}{2c}$ and $y=frac{1}{c}$. Hence
            begin{align}
            2+frac{4c^2-5}{4c}leq 0
            end{align}
            if $c in (-infty, -5/2]$. Hence
            begin{align}
            -frac{5}{2} leq frac{x+2y-2}{x^2+y^2+1}.
            end{align}






            share|cite|improve this answer





















            • +1 Wow nice technique
              – Ovi
              Nov 20 '18 at 5:09
















            3














            For the upper bound observe
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} geq 0
            end{align}
            for all $x, y$ iff
            begin{align}
            2+frac{4c^2-5}{4c}=0 Leftrightarrow c=frac{1}{2}.
            end{align}
            Hence it follows
            begin{align}
            x+2y-2 leq frac{1}{2}(x^2+y^2+1) Longleftrightarrow frac{x+2y-2}{x^2+y^2+1}leq frac{1}{2}
            end{align}
            for all $x, y$. Maximum is attained when $(x, y) = (1, 2)$.



            For the lower bound, we consider
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} leq 0
            end{align}
            which needs to hold for all $x, y$. In particular, the inequality has to hold when $x=frac{1}{2c}$ and $y=frac{1}{c}$. Hence
            begin{align}
            2+frac{4c^2-5}{4c}leq 0
            end{align}
            if $c in (-infty, -5/2]$. Hence
            begin{align}
            -frac{5}{2} leq frac{x+2y-2}{x^2+y^2+1}.
            end{align}






            share|cite|improve this answer





















            • +1 Wow nice technique
              – Ovi
              Nov 20 '18 at 5:09














            3












            3








            3






            For the upper bound observe
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} geq 0
            end{align}
            for all $x, y$ iff
            begin{align}
            2+frac{4c^2-5}{4c}=0 Leftrightarrow c=frac{1}{2}.
            end{align}
            Hence it follows
            begin{align}
            x+2y-2 leq frac{1}{2}(x^2+y^2+1) Longleftrightarrow frac{x+2y-2}{x^2+y^2+1}leq frac{1}{2}
            end{align}
            for all $x, y$. Maximum is attained when $(x, y) = (1, 2)$.



            For the lower bound, we consider
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} leq 0
            end{align}
            which needs to hold for all $x, y$. In particular, the inequality has to hold when $x=frac{1}{2c}$ and $y=frac{1}{c}$. Hence
            begin{align}
            2+frac{4c^2-5}{4c}leq 0
            end{align}
            if $c in (-infty, -5/2]$. Hence
            begin{align}
            -frac{5}{2} leq frac{x+2y-2}{x^2+y^2+1}.
            end{align}






            share|cite|improve this answer












            For the upper bound observe
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} geq 0
            end{align}
            for all $x, y$ iff
            begin{align}
            2+frac{4c^2-5}{4c}=0 Leftrightarrow c=frac{1}{2}.
            end{align}
            Hence it follows
            begin{align}
            x+2y-2 leq frac{1}{2}(x^2+y^2+1) Longleftrightarrow frac{x+2y-2}{x^2+y^2+1}leq frac{1}{2}
            end{align}
            for all $x, y$. Maximum is attained when $(x, y) = (1, 2)$.



            For the lower bound, we consider
            begin{align}
            c(x^2+y^2+1)-x-2y+2 =& cleft(x^2-frac{1}{c}xright)+cleft(y^2-frac{2}{c}yright)+(2+c)\
            =& cleft(x-frac{1}{2c} right)^2+cleft(y-frac{1}{c} right)^2+2+frac{4c^2-5}{4c} leq 0
            end{align}
            which needs to hold for all $x, y$. In particular, the inequality has to hold when $x=frac{1}{2c}$ and $y=frac{1}{c}$. Hence
            begin{align}
            2+frac{4c^2-5}{4c}leq 0
            end{align}
            if $c in (-infty, -5/2]$. Hence
            begin{align}
            -frac{5}{2} leq frac{x+2y-2}{x^2+y^2+1}.
            end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 20 '18 at 5:01









            Jacky Chong

            17.7k21128




            17.7k21128












            • +1 Wow nice technique
              – Ovi
              Nov 20 '18 at 5:09


















            • +1 Wow nice technique
              – Ovi
              Nov 20 '18 at 5:09
















            +1 Wow nice technique
            – Ovi
            Nov 20 '18 at 5:09




            +1 Wow nice technique
            – Ovi
            Nov 20 '18 at 5:09











            0














            How about



            $$left| dfrac {x+2y-2}{x^2+y^2+1} right| le dfrac {|x|+|2y|+|2|}{|x|+|y|+1} le dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$



            This should hold when $|x|, |y| > 1$.



            You can take $K$ to be the unit square. You can check that $fleft(- dfrac 14, - dfrac 12 right) < -2$, so indeed the function has an absolute minimum.






            share|cite|improve this answer




























              0














              How about



              $$left| dfrac {x+2y-2}{x^2+y^2+1} right| le dfrac {|x|+|2y|+|2|}{|x|+|y|+1} le dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$



              This should hold when $|x|, |y| > 1$.



              You can take $K$ to be the unit square. You can check that $fleft(- dfrac 14, - dfrac 12 right) < -2$, so indeed the function has an absolute minimum.






              share|cite|improve this answer


























                0












                0








                0






                How about



                $$left| dfrac {x+2y-2}{x^2+y^2+1} right| le dfrac {|x|+|2y|+|2|}{|x|+|y|+1} le dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$



                This should hold when $|x|, |y| > 1$.



                You can take $K$ to be the unit square. You can check that $fleft(- dfrac 14, - dfrac 12 right) < -2$, so indeed the function has an absolute minimum.






                share|cite|improve this answer














                How about



                $$left| dfrac {x+2y-2}{x^2+y^2+1} right| le dfrac {|x|+|2y|+|2|}{|x|+|y|+1} le dfrac {2|x|+2|y|+2}{|x|+|y|+1} = 2$$



                This should hold when $|x|, |y| > 1$.



                You can take $K$ to be the unit square. You can check that $fleft(- dfrac 14, - dfrac 12 right) < -2$, so indeed the function has an absolute minimum.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 20 '18 at 4:12

























                answered Nov 20 '18 at 4:01









                Ovi

                12.4k1038111




                12.4k1038111






























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