Max and min part 2

Call the expression S ,now using Cauchy-Schwarz Inequality we get $S\leq \sqrt{3(x^2+y^2+z^2)}+x^2+y^2+z^2=1+\sqrt{3}$ The equality holds when $x=y=z=\frac{1}{\sqrt{3}}$

We set $x+y+z=t$ , from the condition we have $t^2-2(xy+yz+xz)=1\Rightarrow xy+yz+xz= \frac{t^2-1}{2}\geq 0$ So $t\geq 1$ , therefore $S=t+\frac{t^2-1}{2}\geq 1$ The equality holds when one variable is equal to 1 and the rest are equal to 0

So finally the sum is $2+\sqrt{3}\approx 3.73$

Maximum of the value follows from,AM-QM inequality and from the fact that $x^2+y^2+z^2 \ge xy+yz+zx$ . For the minimum we see that, the expression equals $\frac{s^2}{2}+s-1$ where s=a+b+c.so,minimizing s would minimize the expression. Since, $x,y,z \in [0,1]$ from the constraint.x+y+z $\ge$ x^2+y^2+z^2=1.Hence ,the minimum is 1/2+1-1/2=1.