What are the possible solutions of the given equation?












2












$begingroup$


I encountered a question in an exam in which we had:




Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






algebra-precalculus







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$endgroup$

















    2












    $begingroup$


    I encountered a question in an exam in which we had:




    Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




    I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






    algebra-precalculus







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I encountered a question in an exam in which we had:




      Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




      I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






      algebra-precalculus







      share|cite|improve this question











      $endgroup$




      I encountered a question in an exam in which we had:




      Find all possible solutions of the equation $$x+y+ {1over x}+{1over y}+4=2 (sqrt {2x+1}+sqrt {2y+1}) $$ where $x $ and $y$ are real numbers.




      I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago









      Thomas Andrews

      130k12147298




      130k12147298










      asked 7 hours ago









      Shashwat1337Shashwat1337

      889




      889






















          2 Answers
          2






          active

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          7












          $begingroup$

          $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



          By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



          $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



          Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            7 hours ago






          • 2




            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            7 hours ago








          • 2




            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            7 hours ago





















          7












          $begingroup$

          It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
          $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
          $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
          $$x=y=1+sqrt2.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hmm, this is the first time I see the cyclic sum notation
            $endgroup$
            – Jan Tojnar
            1 hour ago












          • $begingroup$
            How can we show that the individual summands cannot be non-zero ($x = -y$)?
            $endgroup$
            – Jan Tojnar
            1 hour ago













          Your Answer





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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



          By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



          $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



          Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            7 hours ago






          • 2




            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            7 hours ago








          • 2




            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            7 hours ago


















          7












          $begingroup$

          $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



          By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



          $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



          Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            7 hours ago






          • 2




            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            7 hours ago








          • 2




            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            7 hours ago
















          7












          7








          7





          $begingroup$

          $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



          By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



          $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



          Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$






          share|cite|improve this answer











          $endgroup$



          $$x+y+ {1over x}+{1over y}+4 = x+y+{2x+1over x}+{2y+1over y} $$



          By Am-Gm we have $$ x+{2x+1over x}geq 2sqrt{x{2x+1over x}} = 2sqrt{2x+1}$$ and the same for $y$, so we have



          $$x+y+ {1over x}+{1over y}+4 geq 2sqrt{{2x+1}}+2sqrt{{2y+1}}$$



          Since we have equality is achieved when $x={2x+1over x}$ (and the same for $y$) we have $x=y=1+sqrt{2}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 7 hours ago

























          answered 7 hours ago









          Maria MazurMaria Mazur

          46.9k1260120




          46.9k1260120












          • $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            7 hours ago






          • 2




            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            7 hours ago








          • 2




            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            7 hours ago




















          • $begingroup$
            Truly amazing!!! [+1]
            $endgroup$
            – Dr. Mathva
            7 hours ago






          • 2




            $begingroup$
            @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
            $endgroup$
            – Michael Rozenberg
            7 hours ago








          • 2




            $begingroup$
            AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
            $endgroup$
            – Anurag A
            7 hours ago


















          $begingroup$
          Truly amazing!!! [+1]
          $endgroup$
          – Dr. Mathva
          7 hours ago




          $begingroup$
          Truly amazing!!! [+1]
          $endgroup$
          – Dr. Mathva
          7 hours ago




          2




          2




          $begingroup$
          @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
          $endgroup$
          – Michael Rozenberg
          7 hours ago






          $begingroup$
          @Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
          $endgroup$
          – Michael Rozenberg
          7 hours ago






          2




          2




          $begingroup$
          AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
          $endgroup$
          – Anurag A
          7 hours ago






          $begingroup$
          AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
          $endgroup$
          – Anurag A
          7 hours ago













          7












          $begingroup$

          It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
          $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
          $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
          $$x=y=1+sqrt2.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hmm, this is the first time I see the cyclic sum notation
            $endgroup$
            – Jan Tojnar
            1 hour ago












          • $begingroup$
            How can we show that the individual summands cannot be non-zero ($x = -y$)?
            $endgroup$
            – Jan Tojnar
            1 hour ago


















          7












          $begingroup$

          It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
          $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
          $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
          $$x=y=1+sqrt2.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Hmm, this is the first time I see the cyclic sum notation
            $endgroup$
            – Jan Tojnar
            1 hour ago












          • $begingroup$
            How can we show that the individual summands cannot be non-zero ($x = -y$)?
            $endgroup$
            – Jan Tojnar
            1 hour ago
















          7












          7








          7





          $begingroup$

          It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
          $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
          $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
          $$x=y=1+sqrt2.$$






          share|cite|improve this answer









          $endgroup$



          It's $$sum_{cyc}left(x+frac{1}{x}+2-2sqrt{2x+1}right)=0$$ or
          $$sum_{cyc}frac{x^2-2xsqrt{2x+1}+2x+1}{x}=0$$ or
          $$sum_{cyc}frac{(x-sqrt{2x+1})^2}{x}=0,$$
          which for $xy<0$ gives infinitely many solutions.



          But, for $xy>0$ we obtain:
          $$x=sqrt{2x+1}$$ and $$y=sqrt{2y+1},$$ which gives
          $$x=y=1+sqrt2.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          Michael RozenbergMichael Rozenberg

          108k1895200




          108k1895200












          • $begingroup$
            Hmm, this is the first time I see the cyclic sum notation
            $endgroup$
            – Jan Tojnar
            1 hour ago












          • $begingroup$
            How can we show that the individual summands cannot be non-zero ($x = -y$)?
            $endgroup$
            – Jan Tojnar
            1 hour ago




















          • $begingroup$
            Hmm, this is the first time I see the cyclic sum notation
            $endgroup$
            – Jan Tojnar
            1 hour ago












          • $begingroup$
            How can we show that the individual summands cannot be non-zero ($x = -y$)?
            $endgroup$
            – Jan Tojnar
            1 hour ago


















          $begingroup$
          Hmm, this is the first time I see the cyclic sum notation
          $endgroup$
          – Jan Tojnar
          1 hour ago






          $begingroup$
          Hmm, this is the first time I see the cyclic sum notation
          $endgroup$
          – Jan Tojnar
          1 hour ago














          $begingroup$
          How can we show that the individual summands cannot be non-zero ($x = -y$)?
          $endgroup$
          – Jan Tojnar
          1 hour ago






          $begingroup$
          How can we show that the individual summands cannot be non-zero ($x = -y$)?
          $endgroup$
          – Jan Tojnar
          1 hour ago




















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