Is this simple proof about symmetric sums correct? I'm asked to prove 
$$x \gt 0, y \gt 0, z  \gt 0 \rightarrow$$
$$\left(\frac{x+y}{x+y+z}\right)^\frac{1}{2}+\left(\frac{x+z}{x+y+z}\right)^\frac{1}{2} + \left(\frac{y+z}{x+y+z}\right)^\frac{1}{2} \le 6^\frac{1}{2}$$  
I rewrite the summands and say that it is sufficient to prove:
$$A^\frac{1}{2} + B^\frac{1}{2} + C^\frac{1}{2} \le 6^\frac{1}{2} $$ 
$$ A +B +C = 2$$
$$ 0 \lt A, B, C \le 1$$
Now I just square both sides to get:
$$A + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + B + (BC)^\frac{1}{2} + C \le 6$$
This seems simple:
$$2  + (AB)^\frac{1}{2} + (AC)^\frac{1}{2} + (BC)^\frac{1}{2} \le$$ 
$$ 2  + 1 + 1 + 1 \le 5 \le 6$$  
So I ended up proving that the original bounds were too loose. That makes me worry that I messed up my proof somewhere.

You are almost there.  Squaring both sides yields $$A+B+C+2(AB)^{1/2}+2(BC)^{1/2}+2(CA)^{1/2}\leq 6,$$ and so you only need to prove that $$(AB)^{1/2}+(BC)^{1/2}+(CA)^{1/2}\leq 2.$$  This inequality follows directly from the fact that $A+B+C=2$ along with Cauchy Schwarz.
Alternative Solution: Before squaring both sides, we can deduce desired inequality directly using Cauchy Schwarz.  We have that $$\left(A^{1/2}+B^{1/2}+C^{1/2}\right)^2\leq (A+B+C)(1+1+1),$$ and so we see that $$A^{1/2}+B^{1/2}+C^{1/2}\leq 6^{1/2}.$$
