Jtdylktuy

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}leqsqrt {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 $B$.

Use the fact
$sqrt{(x+y)(x+z)}ge sqrt{xy}+sqrt{xz}$. Which is obvious after squaring and cancelling terms.
Now our inequality becomes :
$$sum limits_{cyc} frac{x}{x+sqrt{(x+y)(x+z)}}le sum limits_{cyc} frac{x}{x+sqrt{xy}+sqrt{xz}}=sum limits_{cyc} dfrac{sqrt{x}}{sqrt{x}+sqrt{y}+sqrt{z}}=1$$

By your work and by C-S we obtain:
$$sum_{cyc}frac{x}{x+sqrt{(x+y)(x+z)}}leqsum_{cyc}frac{x}{2x+sqrt{yz}}=frac{3}{2}-sum_{cyc}left(frac{1}{2}-frac{x}{2x+sqrt{yz}}right)=$$
$$=frac{3}{2}-frac{1}{2}sum_{cyc}frac{sqrt{yz}}{2x+sqrt{yz}}=frac{3}{2}-frac{1}{2}sum_{cyc}frac{yz}{2xsqrt{yz}+yz}leq$$
$$leqfrac{3}{2}-frac{1}{2}cdotfrac{left(sumlimits_{cyc}sqrt{yz}right)^2}{sumlimits_{cyc}(2xsqrt{yz}+yz)}=frac{3}{2}-frac{1}{2}=1.$$