Do I need to plot this?

Let: $u = \sqrt {4z - 1} ,v = \sqrt {4x - 1} ,w = \sqrt {4y - 1}$

Then: $z = \dfrac{{{u^2} + 1}}{4},x = \dfrac{{{v^2} + 1}}{4},y = \dfrac{{{w^2} + 1}}{4}$

Substitute these values into the original system, we have:

$\left\{ \begin{array}{l} {v^2} + {w^2} + 2 = 4u \\ {w^2} + {v^2} + 2 = 4v \\ {u^2} + {v^2} + 2 = 4w \\ \end{array} \right.$

By adding these equations, we get:

$2{u^2} + 2{v^2} + 2{w^2} + 6 = 4u + 4v + 4w$

$\Leftrightarrow 2{(u - 1)^2} + 2{(v - 1)^2} + 2{(w - 1)^2} = 0$

$\Leftrightarrow u = v = w = 1$

$\Leftrightarrow x = y = z = \dfrac{1}{2}$

Hence, $x+y + z = \dfrac{3}{2}$

Adding all three equations,

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

$2x-\sqrt{4x-1}+2y-\sqrt{4y-1}+2z-\sqrt{4z-1}=0$

now, for x (similar for y and z)

2x-√(4x-1)=2[x-√(x-1/4)]

=2[(√(x-1/4))^2+(1/2)^2-2*1/2 √(x-1/4)]

=2(√(x-1/4) - 1/2)^2

Similarly all 3 variables give three squares whose sum is zero

Thus each square is zero.

solving gives each variable as 0.5

Adding them gives 1.5

Method II:

Square all terms and add,

$x^{2}+y^{2}+2xy+y^2+z^{2}+2yz+z^{2}+x^{2}+2xz=4x+4y+4z-1-1-1$

Here,

$x^{2}+y^{2}+1+2xy-2x-2y$

can be replaced by

$(x+y-1)^{2}$

similarly for all other three squares.

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

Since their sum is zero, each of them is zero.

$x+y=1$

$\sqrt{4z-1}=1$

$4z=2$

$z=0.5$

similarly,

x=0.5 and y = 0.5

Their sum is 1.5!!!