Is the question right ?

By Cauchy-Schwarz inequality we know that:

$2(x^2+y^2) \geq (x+y)^2$

Also, $x+y=4-z$ and $x^2+y^2=6-z^2$ , hence:

$2(6-z^2) \geq (4-z)^2$

$12-2z^2 \geq 16-8z+z^2$

$0 \geq 3z^2-8z+4$

$0 \geq (z-2)(3z-2)$

Then, $x$ , $y$ or $z$ must lie in the range $[\frac{2}{3},2]$ , on comparing we get $a=2,b=3,c=2,d=1$ , thus the answer is $2+3+2+1=\boxed{8}$ .

To prove that equality can be achieved:

The discriminant of the polynomial with roots $x$ and $y$ is $(x-y)^2$ , and after a little manipulation we get $(x-y)^2=-3z^2+8z-4$ , and this must be greater than or equal to zero, hence:

$-3z^2+8z-4\geq 0 \\ (z-2)(3z-2) \leq 0 \\ \dfrac{2}{3} \leq z \leq 2$

So, $x$ and $y$ will be real when $z$ is in that range.

At the point where $z$ attains its maximum, we must have $x=y$ , by symmetry; likewise for the minimum of $z$ . Solving $2x+z=4$ and $2x^2+z^2=6$ leads to the equation $3z^2-8z+4=0$ , with solutions $z_{max}=2=\frac{2}{1}$ and $z_{min}=\frac{2}{3}$ , so $a+b+c+d=\boxed{8}$

substituting x in 2nd equation => (4-(y+z)) ^2+y^2+z^2=6

y^2+y(z-4)+z^2-4z+5=0 solving quadratic for y=> y= { (4-z) +/ - sqrt( -3z^2+8z-4) } /2; which gives 3z^2-8z+4 should be >=0; => z lies between {2/3, 2 } for these values y lies in {2/3, 1 } same goes for x; so final range {2/3, 2/1} => a+b+c+d=8;