Proving $\sum_{cyc}\frac{\sqrt{y + z − x}}{\sqrt{y} + \sqrt{z} − \sqrt{x}} \leq 3$ for $x$, $y$, $z$ the sides of a triangle 
Let $x,y$ and $z$ be the length of a triangle. Prove that
$$\frac{\sqrt{y + z − x}}{\sqrt{y} + \sqrt{z} − \sqrt{x}} 
+ \frac{\sqrt{z + x − y}}{\sqrt{z} + \sqrt{x} − \sqrt{y}} 
+ \frac{\sqrt{x + y − z}}{\sqrt{x} + \sqrt{y} − \sqrt{z}} \leq 3$$

I tried to prove that one of the expressions $\leq1$
$x\leq y\leq z$
$∵x-y\leq 0, x-z\leq 0$
$∴(x-y)(x-z) \geq 0$
$x^2-(y+z)x+yz \geq 0$
$x^2+yz \geq x(y+z)$
$x^2+yz \geq xy+xz$
$x^2+yz+2x\sqrt{yz} \geq xy+xz+2x\sqrt{yz}$
$(x+\sqrt{yz})^2 \geq (\sqrt{xy}+\sqrt{xz})^2$
$x+\sqrt{yz} \geq \sqrt{xy}+\sqrt{xz}$
$x \geq \sqrt{xy}+\sqrt{xz}-\sqrt{yz}$
$-2x \leq -2(\sqrt{xy}+\sqrt{xz}-\sqrt{yz})$
$-x \leq -2(\sqrt{xy}+\sqrt{xz}-\sqrt{yz})+x$
$∵y+z > 0$
$y+z-x \leq x+y+z-2(\sqrt{xy}+\sqrt{xz}-\sqrt{yz})$
$y+z-x \leq (\sqrt{y}+\sqrt{z}-\sqrt{x})^2$
$∴ \sqrt{y+z-x} \leq \sqrt{y}+\sqrt{z}-\sqrt{x}$
$\frac{\sqrt{y+z-x}}{\sqrt{y}+\sqrt{z}-\sqrt{x}} \leq 1$
Similarly, the other two expressions can also be proven the same way.
Therefore
$$\frac{\sqrt{ +  − }}{\sqrt{} + \sqrt{} − \sqrt{}} + 
\frac{\sqrt{ +  − }}{\sqrt{} + \sqrt{} − \sqrt{}} +
\frac{\sqrt{ +  − }}{\sqrt{} + \sqrt{} − \sqrt{}} \leq 3$$
Is my proof correct？
$$\frac{\sqrt{ +  − }}{\sqrt{} + \sqrt{} − \sqrt{}} \geq 1$$
and
$$\frac{\sqrt{ +  − }}{\sqrt{} + \sqrt{} − \sqrt{}} \leq 1$$
My proof is incomplete because the 2nd term does not sastify my proof. May I know what is the next steps?

For the second term, when we'll do the proving we'll shake hands with the expression $$z+x-y \leq (\sqrt{z}+\sqrt{x}-\sqrt{y})^2$$ This is not true for the triplet $(n-1,n,n+1)$
I'll leave it for the OP to answer the reason for this.
