3
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Hmm, this is the first time I see the cyclic sum notation $\endgroup$ – Jan Tojnar Mar 15 at 1:00
  • $\begingroup$ How can we show that the individual summands cannot be non-zero ($x = -y$)? $\endgroup$ – Jan Tojnar Mar 15 at 1:12
  • $\begingroup$ Thank you Michael! $\endgroup$ – Shashwat1337 Mar 15 at 7:40
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Mar 15 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 at 7:51
7
$\begingroup$

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 $x=y=1+\sqrt{2}$

$\endgroup$
  • 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 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 at 7:46
  • $\begingroup$ @Maria Mazur After your fixing I deleted my previous comment, Now your statement is true. $\endgroup$ – Michael Rozenberg Mar 15 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.