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

Find all possible solutions of the equation $$x+y+ {1\over x}+{1\over y}+4=2 (\sqrt {2x+1}+\sqrt {2y+1}) $$ 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?


It's $$\sum_{cyc}\left(x+\frac{1}{x}+2-2\sqrt{2x+1}\right)=0$$ or $$\sum_{cyc}\frac{x^2-2x\sqrt{2x+1}+2x+1}{x}=0$$ 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 $$x=y=1+\sqrt2.$$

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  • $\begingroup$ Hmm, this is the first time I see the cyclic sum notation $\endgroup$ – Jan Tojnar Mar 15 '19 at 1:00
  • $\begingroup$ How can we show that the individual summands cannot be non-zero ($x = -y$)? $\endgroup$ – Jan Tojnar Mar 15 '19 at 1:12
  • $\begingroup$ Thank you Michael! $\endgroup$ – Shashwat1337 Mar 15 '19 at 7:40
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Mar 15 '19 at 7:41
  • $\begingroup$ @Jan Tojnar If $y=-x$ we obtain $1=\sqrt{1-4x^2}$ or $x=0,$ which is impossible. $\endgroup$ – Michael Rozenberg Mar 15 '19 at 7:51

This solution works only for positive $x,y$. However they can not be both negative since then LHS is at most $0$. So $$x+y+ {1\over x}+{1\over y}+4 = x+y+{2x+1\over x}+{2y+1\over y} $$

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

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

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

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  • 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 Mar 14 '19 at 19:04
  • $\begingroup$ @Maria Mazur Yes, it appears to be wrong as AM-GM only applies to positive real numbers. But still, thanks for providing me with an alternative solution for positive numbers. $\endgroup$ – Shashwat1337 Mar 15 '19 at 7:46
  • $\begingroup$ @Maria Mazur After your fixing I deleted my previous comment, Now your statement is true. $\endgroup$ – Michael Rozenberg Mar 15 '19 at 13:23

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