Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$

Question

Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$

First natural move would be rewriting the system as: $$x^{2}+y^{2}+2xy=4z-1$$ $$x^{2}+z^{2}+2xz=4y-1$$ $$z^{2}+y^{2}+2zy=4x-1$$ Thus, $$2x^{2}+2y^{2}+2z^{2}+2xy+2yz+2zx-4x-4y-4z+3=0$$ And I tried to write the polynomial above as the sum of complete squares. Is this possible or do you have other ideas?

Answer 1

Note that these equations are defined only for $x,y,z\ge 1/4$.

Consider the first two equations $$x+y=\sqrt{4z-1}$$ $$x+z=\sqrt{4y-1}$$ Subtracting these two equations gives us $$y-z=\sqrt{4z-1}-\sqrt{4y-1}$$ If $y\gt z$, then the LHS is positive but the RHS is negative, so this cannot be. Similarly, if $y\lt z$, then the LHS is negative but the RHS is positive, which is also impossible. Thus, we have that $$y-z=\sqrt{4z-1}-\sqrt{4y-1}=0$$ or $$y=z$$ We may show by similar reasoning that $x=y$, and so $x=y=z$. Then we have that $$2x=\sqrt{4x-1}$$ or $$4x^2-4x+1=0$$ Which has one solution: $$x=\frac{1}{2}$$ This gives us the only solution to your system: $$x=y=z=\frac{1}{2}$$

NOTE: This problem is trivialized by the fact that both $x$ and $\sqrt{4x-1}$ are increasing functions. The following system of equations might be more interesting: $$x+y=\frac{1}{\sqrt{4z-1}}$$ $$x+z=\frac{1}{\sqrt{4y-1}}$$ $$y+z=\frac{1}{\sqrt{4x-1}}$$

Answer 2

Solution. $\blacktriangleleft$ Subtract every pair of the equation, we would obtain something like \begin{align*} y-z &= (x+y) - (x+z) \\ &= \sqrt {4z-1} - \sqrt {4y-1} \\&= \frac {4z-1 - 4y+1} {\sqrt {4z-1} + \sqrt {4y-1}} \\&= 4\frac {z-y} {\sqrt {4z-1} + \sqrt {4y-1}}, \end{align*} then $y -z$ must be zero, otherwise a sum of two square roots would be negative, which is impossible. Thus we have $x = y =z$. Solve any one of them would obtain that $$ x = y = z = \frac 12. \blacktriangleright $$

Answer 3

Necessarily $x,y,z\ge{\large{\frac{1}{4}}}$.

Without loss of generality, assume $x=\min(x,y,z)$.

Then $2x \le y+z$, hence \begin{align*} &(y+z)^2=4x-1\\[4pt] \implies\;&(2x)^2\le 4x-1\\[4pt] \implies\;&4x^2-4x+1\le 0\\[4pt] \implies\;&(2x-1)^2\le 0\\[4pt] \implies\;&(2x-1)^2=0\\[4pt] \implies\;&x={\small{\frac{1}{2}}}\\[4pt] \end{align*} Then $y,z\ge {\large{\frac{1}{2}}}$, and $4x-1=1$, hence \begin{align*} &(y+z)^2=4x-1\\[4pt] \implies\;&(y+z)^2=1\\[4pt] \implies\;&y+z=1\\[4pt] \implies\;&y=z={\small{\frac{1}{2}}}\\[4pt] \end{align*} Therefore $x=y=z={\large{\frac{1}{2}}}$.