Non-homogeneous cyclic $\frac{x+1}{\sqrt{x+y}}+\frac{y+1}{\sqrt{y+z}}+\frac{z+1}{\sqrt{z+x}} \geq 3\sqrt{2}$ 
Consider three non-negative real numbers $x$, $y$ and $z$, no two of
  which are zero at the same time. Prove that the following inequality
  holds:
$$\frac{x+1}{\sqrt{x+y}}+\frac{y+1}{\sqrt{y+z}}+\frac{z+1}{\sqrt{z+x}} \geq 3\sqrt{2}$$

My progress: This inequality is cyclic, so we should consider two possibilities $x \geq y\geq z$ and $x \leq y\leq z$. The first case is relatively simple because from AM-GM:
$$\sum \frac{x+1}{\sqrt{2(x+y)}} \geq  \sum \frac{2(x+1)}{x+y+2}$$
and if we let $a = x+1,\ b=y+1,\ c= z+1$, ($a\geq b\geq c$), we have:
$$2\sum \frac{a}{a+b}-3 = \frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)} \geq 0$$
However, I believe the second case ($x\leq y \leq z$) is very difficult to prove, and I am not sure if this is the way to proceed.
 A: Remarks: The inequality (1) is @arqady's idea. Although the original inequality is not homogeneous, (1) is. Hope to see a nice proof of (1).
A proof utilizing arqady@AoPS's nice idea.
Since the desired inequality is cyclic,
assume that $z = \min(x, y, z)$.
The desired inequality is written as
$$\sum_{\mathrm{cyc}} \frac{x}{\sqrt{x + y}} + \sum_{\mathrm{cyc}} \frac{1}{\sqrt{x + y}} \ge 3\sqrt 2.$$
Squaring both sides, it suffices to prove that
$$\left(\sum_{\mathrm{cyc}} \frac{x}{\sqrt{x + y}} + \sum_{\mathrm{cyc}} \frac{1}{\sqrt{x + y}}\right)^2 \ge 18.$$
Using $(A + B)^2 \ge 4AB$, it suffices to prove that
$$4 \sum_{\mathrm{cyc}} \frac{x}{\sqrt{x + y}} \cdot \sum_{\mathrm{cyc}} \frac{1}{\sqrt{x + y}} \ge 18$$
or (simplifying)
$$\sum_{\mathrm{cyc}} \frac{x}{x + y} + \sum_{\mathrm{cyc}} \sqrt{\frac{x + y}{y + z}} \ge \frac{9}{2} \tag{1}.$$
(1) is written as
$$\sum_{\mathrm{cyc}} \sqrt{\frac{x + y}{y + z}}\ge 3 + \frac{(y - x)(x - z)(y - z)}{2(x + y)(y + z)(z + x)}.$$
Using AM-GM, we have $\mathrm{LHS} \ge 3$.
Thus, we only need to prove the case when $y \ge x \ge z$.
Using $\sqrt{u} \ge \frac{1 + 3u}{3 + u}$ for all $u \ge 1$, we have
\begin{align*}
 \sqrt{\frac{x + y}{y + z}} &\ge \frac{3x + 4y + z}{x + 4y + 3z}, \\
 \sqrt{\frac{y + z}{z + x}} &\ge
 \frac{x + 3y + 4z}{3x + y + 4z}.
\end{align*}
Also, we have
$$\sqrt{\frac{z + x}{x + y}}
= \left(\sqrt{\frac{x + y}{y + z}\cdot\frac{y + z}{z + x}}\right)^{-1} 
\ge \left(\frac{\frac{x + y}{y + z} + \frac{y + z}{z + x}}{2}\right)^{-1}.$$
Thus, it suffices to prove that
$$\frac{3x + 4y + z}{x + 4y + 3z}
+ \frac{x + 3y + 4z}{3x + y + 4z} + \left(\frac{\frac{x + y}{y + z} + \frac{y + z}{z + x}}{2}\right)^{-1} \ge 3 + \frac{(y - x)(x - z)(y - z)}{2(x + y)(y + z)(z + x)}.$$
Letting $x = z + s, y = z + s + t$ for $s, t \ge 0$, we have
$$\mathrm{LHS} - \mathrm{RHS} = \frac{f(z, s, t)}{g(z, s, t)}$$
where $f(z, s, t)$ and $g(z, s, t)$ are both polynomials
with non-negative coefficients. The inequality is true.
We are done.
