Is this problem wrongly built? Or is there a solution which I don't know how to arrive at? I was solving a Cauchy-Schwarz's inequality based problem. Given that $x^2+y^2+z^2=1$ I am supposed to show that $x+y+z \le 6$. After struggling for a while I realised that I could solve this inequality had the condition been $x^2+y^2+z^2=12$. This was my solution:
$(x.1+y.1+z.1)^2 \le (x^2+y^2+z^2)(1^2+1^2+1^2)$
$\implies (x+y+z)^2 \le 12*3$
$\implies (x+y+z)^2 \le 12*3 \le 36$
Taking square root on both sides, I got
$(x+y+z) \le 6$
Now I am wondering whether I detected a typo in the problem ($x^2+y^2+z^2=12$, not $x^2+y^2+z^2=1$) or is there a real way to reach $(x+y+z) \le 6$ starting from $x^2+y^2+z^2=1$ based on Cauchy-Schwarz principles! 
The main reason for this doubt is I am stuck on an extension of this problem in which I need to prove $x^3+y^3+z^3 \ge 24$
Any help is appreciated. 

It seems that there is no need for Cauchy inequality. Obviously $x\leq1$. (why?). Similarly $y$ and $z$ are. So $x+y+z\leq 3$. This inequality can be strengthen since x, y, z can't take value 1 at the same time.
