inequality - Prove $sumlimits_{cyc} frac{sqrt{xy}}{sqrt{xy+z}}lefrac{3}{2}$ if $x+y+z=1$

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.

Answer

$$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2} $$
$$ \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+1-x-y}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(1-x)(1-z)}} = \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(y+z)(x+z)}} $$
And now, by AM-GM
$$ \sum_{cyc} \frac{\sqrt{xy}}{\sqrt{(y+z)(x+z)}} \le \sum_{cyc} \frac{x}{2(x+z)}+\frac{y}{2(y+z)} = \frac{x}{2(x+z)}+\frac{y}{2(y+z)} + \frac{y}{2(y+x)}+\frac{z}{2(z+x)}+ \frac{z}{2(z+x)}+\frac{x}{2(x+y)} = \frac{3}{2} \Box $$