Find real solutions to a system

[anemone]

Suppose that $x$ and $y$ are positive real numbers. Find all real solutions of the equation $\dfrac{2xy}{x+y}+\sqrt{\dfrac{x^2+y^2}{2}}=\sqrt{xy}+\dfrac{x+y}{2}$.

 

[topsquark]

[sp]
As usual with me I can't find an efficient (or in this case even elegant) way to do this. But I had an idea...

[math]\dfrac{2xy}{x + y} + \sqrt{ \dfrac{x^2 + y^2}{2} } = \sqrt{xy} + \dfrac{x + y}{2}[/math]

Let a = y/x.
Two points:
1) Watch out for x = 0 later.
2) I tacitly multiplied the first fraction by a/a. So watch out of a = 0, as well.

Then, after a bit
[math]\dfrac{2a}{a + 1} + \dfrac{1}{ \sqrt{2} } ~ \sqrt{a^2 + 1} = \sqrt{a} + \dfrac{1}{2} (a + 1)[/math]

Isolate the [math]\sqrt{a + 1}[/math] and square. After some more simplifying and clearing the fractions
[math]2 (a^2 + 1)(a + 1)^2 = 16 a^2 + 4 a(a + 1)^2 + (a + 1)^4 - 8 a (a + 1)^2 + ( -16 a(a + 1) + 4 a (a + 1)^3 ) \sqrt{a}[/math]

Expanding and simplifying
[math]2a^4 + 4a^3 + 4a^2 + 4a + 2 = (a^4 + 14a^2 + 1) + (4a^3 - 4a^2 - 4a + 4) \sqrt{a}[/math]

[math]a^4+ 4a^3 - 10a^2 + 4a + 1 = (4a^3 - 4a^2 - 4a + 4) \sqrt{a}[/math]

So far my approach for finding a is pretty standard. Isolate one square root, simplify, then the next step would be to isolate the other square root and simplify. But there is an issue with this procedure: If a = 1 (as it will be when we solve it) then we have both sides of this equal to 0! I have chosen to change the variable to [math]b^2 = a[/math]. This means we don't have to square out that LHS, which is ugly enough. But again, we have to be careful about values of b now. (Don't worry, it'll all come together.)

So with the substitution and (yet more) simplifying:
[math]b^8 - 4 b^7 + 4 b^6 + 4 b^5 - 10 b^4 + 4 b^3 + 4 b^2 - 4 b + 1 = 0[/math]

The rational root theorem says that [math]b = \pm 1[/math] are the only rational solutions. Instead of some fancy theorems to prove these are the only real solutions, I simply graphed it. It was faster.

So, [math]b = \pm 1[/math] are the only real solutions. Time to work backwards. [math]a = b^2 = 1[/math]. Thus [math]\dfrac{y}{x} = 1[/math] and we get that the solutions to the original equation are all y = x. What about the x = 0 thing? y = x = 0 violates the original equation.

So the final solution is all real [math]y = x, ~ x \neq 0[/math].
[/sp]

-Dan

 

[anemone]

Spoiler

Well done, topsquark! You know, I sense your obsession lately with my POTWs and challenge problems, hehehe...I hope so far you found nothing but fun in tackling all those problems!

 

[topsquark]

Spoiler

Well done, topsquark! You know, I sense your obsession lately with my POTWs and challenge problems, hehehe...I hope so far you found nothing but fun in tackling all those problems!

[sp]
It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.
[/sp]

-Dan

 

[MarkFL]

[sp]
It's not so much that I've been obsessed, it's more that I have actually been able to solve some recently. I usually try to work them out.
[/sp]

-Dan

Spoiler

 

[topsquark]

[sp](Whispers) Why are we having a conversation using spoilers?[/sp]

-Dan

 

[anemone]

Spoiler

Because it is fun! (Bow)(Emo)(Drink)