Prove $\sum\limits_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le\frac{3}{2}$ if $x+y+z=1$

Question

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 1

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

Answer 2

HINT 1: $$\sum_{cyc} \frac{\sqrt{xy}}{\sqrt{xy+z}}\le \sqrt{\left(\sum_{cyc}xy\right)\left(\sum_{cyc}\frac1{xy+z}\right)}$$

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HINT 2: $$\frac{\sqrt{xy}}{\sqrt{xy+z}}=\frac{1}{\sqrt{1+\frac{z}{xy}}}$$ Let $a=z/xy,b=x/yz,a=y/zx$, now $abc=1$, so $(1+a)\ge a$ So $$\sum_{cyc}\frac1{\sqrt{1+a}}\le\sum_{cyc}\frac1{\sqrt{a}}=\sum_{cyc}\frac{\sqrt{xy}}{\sqrt z}$$