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Let $(x,y,z)$ be the real solution to the system of equations \begin{align*} x+y &= \sqrt{4z-1} ,\\ y+z &= \sqrt{4x- 1} , \\ z+x &= \sqrt{4y-1} . \end{align*} Find $x+y+z.$

I could add all the equations up but that doesn't do any good. Thanks in advance!

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    $\begingroup$ First square them, then add them. Also, do you know what you get when you expand $(x+y+z)^2$? $\endgroup$ – Arthur Aug 26 '16 at 22:22
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$$x+y=\sqrt{4z-1}\tag{1}$$

$$y+z=\sqrt{4x-1}\tag{2}$$

$$z+x=\sqrt{4y-1}\tag{3}$$

Adding $(1),(2)$ and $(3)$, we get

$$x+y+z=\sqrt{x-\frac14}+\sqrt{y-\frac14}+\sqrt{z-\frac14}$$

Let $a=\sqrt{x-\frac14}$,$b=\sqrt{y-\frac14}$ and $c=\sqrt{z-\frac14}$

Using these substitutions, we get

$$\left(a^2+\frac14\right)+\left(b^2+\frac14\right)+\left(c^2+\frac14\right)=a+b+c$$ $$ \Longleftrightarrow \left(a-\frac12\right)^2+\left(b-\frac12\right)^2+\left(c-\frac12\right)^2=0$$

Thus, the only possible solution is $\color{blue}{\boxed{\color{red}{(x,y,z)=\left(\frac12,\frac12,\frac12\right)}}}$

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  • $\begingroup$ Perfect! Thanks for answering, Roby5! $\endgroup$ – Dreamer Aug 27 '16 at 0:25
  • $\begingroup$ This is a perfect example of how powerful the inequality $x^2 \geq 0$ is!! +1 $\endgroup$ – Airdish Aug 28 '16 at 10:53
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To initiate Arthur's hints in the comments: $$x^2 + 2xy + y^2 = (x + y)^2 = 4z - 1$$ $$y^2 + 2yz + z^2 = (y + z)^2 = 4x - 1$$ $$z^2 + 2zx + x^2 = (z + x)^2 = 4y - 1$$

Rather than adding all three equations, we subtract them pairwise. Then assuming $x \neq y \neq z$, we have: $$2y(x - z) + (x + z)(x - z) = 4(z - x) \implies 2y + x + z = -4$$ $$2z(y - x) + (y + x)(y - x) = 4(x - y) \implies 2z + y + x = -4$$ $$2x(y - z) + (y + z)(y - z) = 4(z - y) \implies 2x + y + z = -4$$

Equating, we get $$2y + x + z = 2z + y + x = 2x + y + z = -4.$$

Therefore, we obtain $$x + y + z = -4 - y = -4 - z = -4 - x.$$

This implies that $$x = y = z,$$ which contradicts our earlier assumption.

Therefore, the original system of equations has no real solution.

Added August 27 2016

Therefore, for the original system of equations to have a real solution, we must have $x = y = z$.

Consequently, the system is reduced to solving the lone equation $$x + x = \sqrt{4x - 1}$$ $$(2x)^2 = 4x - 1$$ $$4x^2 - 4x + 1 = 0$$ $$x = \frac{1}{2}.$$

Thus, $$x = y = z = \frac{1}{2}$$ is the only real solution.

QED

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  • $\begingroup$ I concede that Roby5's answer is more elegant (and correct!). However, I leave this answer undeleted (and will henceforth flag it as community wiki), in the hope that somebody will be able to mend it. $\endgroup$ – Jose Arnaldo Bebita-Dris Aug 26 '16 at 22:52
  • $\begingroup$ Answer has been clarified (and subsequently edited). $\endgroup$ – Jose Arnaldo Bebita-Dris Aug 26 '16 at 22:57
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We can solve the system for $x, y, z$. By the symmetry of the equations, we can say $x=y=z$, and solve any of them for this value. (all equations become the same one under that conditions: $2t = \sqrt{4t-1}$) We would get $t= \frac12$ for $t = x=y= z$.

So, we found a faster way to one of the solutions. It is not hard to prove that it is also the only solution. For example, by squaring the first two equations and after subtraction:

$(x+y)^2 - (y+z)^2 = 4 (z-x)$

If $z>x$ then the right side is positive, and the left negative. If $z<x$ then the right side is negative, and the left positive. So $x=z$. The same way we can do it with the second and third equation to get $x=y$. (it is also clear that $x, y, z$ must be positive numbers so the roots are defined)

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  • $\begingroup$ Symmetry of equation does not automatically mean symmetry of solution (say $x+y=1,x^2+y^2=1$), although it suggests that that's an easy place to start looking. $\endgroup$ – Arthur Aug 26 '16 at 22:33
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    $\begingroup$ By the same symmetry argument, the system $x + y = 1$ and $x^2 + y^2 = 1$ would have a solution $x = y$ which is obviously false. $\endgroup$ – dxiv Aug 26 '16 at 22:34
  • $\begingroup$ I edited my answer to clarify how we can prove that all variables have to be equal. I didn't mean to say that symmetry is an argument, but it sugests something: in this case all equations would be equivalent when $x=y=z$. So I concluded that it has to be the solution, even witouth solving it. The rest is to show that it is the only solution. $\endgroup$ – Icarus 369 Aug 26 '16 at 22:42
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@Roby5 @user11235813

This question can benefit from a CAS (Computer Algebra System) assistance.

As this kind of tools is still not widespread, I give here the Mathematica requests I have used (other CAS work along very similar lines)

g = GroebnerBasis[{$4z - 1 - (x + y)^2, 4x - 1 - (y + z)^2, 4y - 1 - (z + x)^2$}]

p = g[[1]]

Factor[p]

Answer:

$\{ 1241 - 4156\,z + 2044\,z^2 + 2016\,z^3 + 1136\,z^4 + 64\,z^5 + 64\,z^6,$

$ 17 - 918\,y + 879\,z - 648\,y\,z + 616\,z^2 - 216\,y\,z^2 + 272\,z^3 + 48\,z^4 + 16\,z^5,$

$-493 + 10368\,y + 2592\,y^2 - 8778\,z + 5184\,y\,z - 8360\,z^2 - 976\,z^3 - 528\,z^4 - 32\,z^5,$

$-3085 + 10368\,x + 10368\,y - 8778\,z - 10952\,z^2 - 976\,z^3 - 528\,z^4 - 32\,z^5\}$

$1241 - 4156\,z + 2044\,z^2 + 2016\,z^3 + 1136\,z^4 + 64\,z^5 + 64\,z^6$

${\left( -1 + 2\,z \right) }^2\,\left( 73 - 4\,z + 4\,z^2 \right) \, \left( 17 + 12\,z + 4\,z^2 \right)$

Explanations : we have chosen the first polynomial in a list of 4, because it is the only one which is expressed with a single variable ($z$).

What conclusion can be drawn ?

That, necessarily, unknown $z$ should verify:

$$p(z):= 1241 - 4156\,z + 2044\,z^2 + 2016\,z^3 + 1136\,z^4 + 64\,z^5 + 64\,z^6=0$$

which can be given under the following factorized form:

$$p(z)=\left( -1 + 2\,z \right)^2\,\left( 73 - 4\,z + 4\,z^2 \right) \, \left( 17 + 12\,z + 4\,z^2 \right)=0$$

where quadratic factors have no real roots.

Thus, necessarily $z=\dfrac{1}{2}$.

Due to the circularity of the given system, $x=y=\dfrac{1}{2}$ as well.

(immediate checking : $x=y=z=\dfrac{1}{2}$ is a solution...).

Remark: I don't give here any explanation about Groebner bases. IMHO, it is a kind of tool the existence of which should be known (the same is true for other algebraic tools like resultants, Sturm sequences, etc.) ; a deep understanding of the background theories is not necessary beyond a certain point in a first step. But most people having seen the power of these tools, their mathematical appetite being wetted, will sooner or later desire to understand "what is under the hood"...

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