prove $\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$ 
If $x$,$y$,$z$ are positive real numbers,Prove:$$\sum \limits_{cyc} \frac{x}{x+\sqrt{(x+y)(x+z)}}\leq 1$$ 
  Using this two inequality:
$\sum ^n_{i=1} \sqrt{a_ib_i}\leq\sqrt {ab} $ (we call it $A$ inequality)
$\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$ (we call it $B$ inequality)
which $a_i$ and $b_i$ are positive and and $b= \sum ^n_{i=1} b_i$,$a= \sum ^n_{i=1} a_i$.
Additional info: The question emphasizes in using inequalities $A$ and $B$.Beside them we can use AM-GM and Cauchy inequalities only.We are not allowed to use induction.And if you like,here you can see Prove of inequality B.

Things I have tried so far: 
Using inequality $A$ i can re write question inequality as$$\sum \limits_{cyc} \frac{x}{2x+\sqrt{xy}}\leq 1$$
And I can't go further.I can't observer something that could lead me to using inequality $B$.

By your work and by C-S we obtain:
$$\sum_{cyc}\frac{x}{x+\sqrt{(x+y)(x+z)}}\leq\sum_{cyc}\frac{x}{2x+\sqrt{yz}}=\frac{3}{2}-\sum_{cyc}\left(\frac{1}{2}-\frac{x}{2x+\sqrt{yz}}\right)=$$
$$=\frac{3}{2}-\frac{1}{2}\sum_{cyc}\frac{\sqrt{yz}}{2x+\sqrt{yz}}=\frac{3}{2}-\frac{1}{2}\sum_{cyc}\frac{yz}{2x\sqrt{yz}+yz}\leq$$
$$\leq\frac{3}{2}-\frac{1}{2}\cdot\frac{\left(\sum\limits_{cyc}\sqrt{yz}\right)^2}{\sum\limits_{cyc}(2x\sqrt{yz}+yz)}=\frac{3}{2}-\frac{1}{2}=1.$$
