if $x,y,z$ are positive real numbers and $x+y+z=1$ Prove:$$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$$ where $\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $x,y,z$.
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.
Things I have done so far: Using AM-GM $$xy+z \ge 2$$ $$\sqrt{xy+z} \ge \sqrt2$$ So manipulating this leads to $$\sum_{cyc}\frac{\sqrt{xy}}{\sqrt{xy+z}} \le \sum_{cyc}\frac{\sqrt{xy}}{\sqrt2}$$
I stuck here.I'm thinking about applying Cauchy-Schwartz.Also I have not used the assumption $x+y+z=1$.Any hint is appreciated.