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lemon14 · Oct 5, 2010, 06:53 AM

$\sqrt[1+2x-y^2]{\sqrt[1+2y-z^2]{\sqrt[1+2z-x^2]{x+y+z}}}, x,y,z\epsilon N \Longrightarrow x,y,z = ?$

I have no idea how to get started. I fancy that I don't have to calculate all the arguments :-/

Unknown008 · Oct 5, 2010, 07:31 AM

$\sqrt[1+2x-y^2]{\sqrt[1+2y-z^2]{\sqrt[1+2z-x^2]{x+y+z}}} = (x+y+z)^{(\frac{1}{1+2z-x^2})(\frac{1}{1+2y-z^2})(\frac{1}{1+2x-y^2})}$

Given $x,\ y,\ z\ \in\ N\ \Longrightarrow\ x,\ y,\ z = ?$

Is that it?

lemon14 · Oct 5, 2010, 07:35 AM

I'm sorry, I don't know how much it should equal, I only know that I have to find out the values for x,y,z. How did you get this?

Unknown008 · Oct 5, 2010, 07:43 AM

In the same way that we know that:

$\sqrt{x} = x^{\frac12}$

If we had $\sqrt{\sqrt{\sqrt{x}}} = \sqrt{\sqrt{x^{\frac12}}} = \sqrt{x^{(\frac12)^2}} = x^{(\frac12)^3}$

lemon14 · Oct 5, 2010, 08:00 AM

Ok, then it means that $1+2x-y^2=1+2y-z^1=1+2z-x^2$ ?

Unknown008 · Oct 5, 2010, 08:06 AM

Oh shoo sorry sorry sorry :(

I didn't notice that the roots were different in each case :(

I'm editing my post.

Okay, I corrected my previous mistake and it seems that you need to expand.

But the thing is, I never encountered such a problem and I don't know how to solve it :(

I just wanted to clarify what you were asking.

I don't know if it's right, but that's what I think:

To find the roots, we need to equate the expression to zero.

So, we have:

$(x+y+z)^{(\frac{1}{1+2z-x^2})(\frac{1}{1+2y-z^2})(\frac{1}{1+2x-y^2})} = 0$

But whatever be the power of (x+y+z), it'll be greater than zero if $x+y+z \neq 0$

So, we need to find when $x + y + z = 0$

But for that... there are infinite solutions :(

EDIT: "it'll be greater than zero if $x+y+z \neq 0$ " considering $|x+y+z|$

lemon14 · Oct 5, 2010, 09:01 AM

Ok, I'll let you know if I figure out how to solve it.

lemon14 · Oct 6, 2010, 09:34 AM

I was thinking what if we would write: $\{1+2x-y^2>1\\1+2y-z^2>1\\1+2z-x^2>1$ , then sum them up: $3+2(x+y+z)-(x^2+y^2+z^2)>3$
$(x^2+y^2+z^2)-2(x+y+z)<0$ ? But from now on I don't know how to find the values :(

lemon14 · Oct 9, 2010, 04:43 AM

the solution:
$\{1+2x-y^2\geq 2\\ 1+2y-z^2\geq 2\\ 1+2z-x^2\geq 2$
$y^2-2x+1\leq 0\\ z^2-2y+1\leq 0\\ x^2-2z+1\leq 0$
$(x^2-2x+1)+(y^2-2y+1)+(z^2-2z+1)\leq 0$
$(x-1)^2+(y-1)^2+(z-1)^2\leq 0$
$(x-1)^2+(y-1)^2+(z-1)^2$ cannot be <0 so:
$(x-1)^2+(y-1)^2+(z-1)^2=0$
$\Rightarrow\{x=1\\y=1\\z=1$

Unknown008 · Oct 9, 2010, 06:02 AM

Wow, I never saw that way of solving...

And I don't even understand how you got greater than or equal to 2... I guess I'll learn it some other time :)

lemon14 · Oct 10, 2010, 03:50 AM

It is not that difficult: the littlest degree of a root is 2 and here the degrees can be either 2 or greater, they cannot be 1 or smaller. Then I put them into a system of inequalities and made some calculations. The problem wouldn't have been so difficult if I thought of this little rule. I'm sorry if I am not very clear, but I don't have the specific vocabulary to explain it as I'd like to.