# What are the possible solutions of $x+y+ {1\over x}+{1\over y}+4=2 (\sqrt {2x+1}+\sqrt {2y+1})$?

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 $$x=y=1+\sqrt2.$$

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

• AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative. – Anurag A Mar 14 at 19:04
• @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. – Shashwat1337 Mar 15 at 7:46
• @Maria Mazur After your fixing I deleted my previous comment, Now your statement is true. – Michael Rozenberg Mar 15 at 13:23