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. 
 A: In the first part of your problem, assuming $x^2+y^2+z^2=1$ is actually no problem, because $x^2+y^2+z^2=1$ implies $x,y,z\le 1$, so $x+y+z\le 3<6$. 
Also, using Cauchy-Schwarz you can slightly improve this bound to $x+y+z\le \sqrt{3}$.
Looking however also at your next problem, showing that $x^3+y^3+z^3\ge 24$, it seems that you are right. The assumption $x^2+y^2+z^2=1$ does not imply this, because clearly $1^2+0^2+0^2=1$, but $1^3+0^3+0^3=1<24$.
Taking however, your modified assumption $x^2+y^2+z^2=12$ we obtain by Cauchy-Schwarz:
$$144=(x^\frac32 x^\frac12+y^\frac32 y^\frac12+z^\frac32 z^\frac12)^2 \le (x^3+y^3+z^3)(x+y+z)\le 6(x^3+y^3+z^3),$$
as required. Therefore, you are right in saying, that this is most likely a typo in the problem statement.
A: 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.
A: We have $x^2\le1$, and thus $x\le1$. Similarly, we have $x+y+z\le3<6$.
