Find $x+y+z$ From Equations Including Radicals

Question

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!

Answer 1

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

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

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

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 $\color{blue}{\boxed{\color{red}{(x,y,z)=\left(\frac12,\frac12,\frac12\right)}}}$

Answer 2

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)

Answer 3

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

Answer 4

@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"...