Let $f: \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that:

1. $f(xyz) + f(x) + f(y) + f(z) = f(\sqrt{xy}) f(\sqrt{yz}) f(\sqrt{zx})$ for all $x, y, z \in \mathbb{R}^+$.
2. $f(x) < f(y)$ for all $1 \le x < y$.

We are tasked with finding all such functions $f$.

### Step 1: Analyze the Symmetry in the Functional Equation

The given functional equation is symmetric in $x, y, z$. Hence, we try to find simple forms of $f(x)$ by testing functions that exhibit symmetry.

### Step 2: Consider Simple Forms

Suppose $f(x) = x^k$ for some exponent $k$. Then substituting into the functional equation, we have:
$$(xyz)^k + x^k + y^k + z^k = (\sqrt{xy})^k (\sqrt{yz})^k (\sqrt{zx})^k.$$

The right-hand side simplifies to:
$$(xy)^{\frac{k}{2}} (yz)^{\frac{k}{2}} (zx)^{\frac{k}{2}} = (xyz)^k.$$

Thus, to maintain equality, the additional terms $x^k + y^k + z^k$ suggest considering functions of the form $f(x) = x^k + \frac{1}{x^k}$.

### Step 3: Verify the Conditions

Let's verify $f(x) = x^k + \frac{1}{x^k}$ against the functional equation. Plug this form in for $f$:
$$f(xyz) = (xyz)^k + \frac{1}{(xyz)^k}$$
and
$$f(\sqrt{xy}) = (xy)^{\frac{k}{2}} + \frac{1}{(xy)^{\frac{k}{2}}}.$$

For the equation:
$$(xyz)^k + \frac{1}{(xyz)^k} + x^k + \frac{1}{x^k} + y^k + \frac{1}{y^k} + z^k + \frac{1}{z^k}$$
$$= \left((xy)^{\frac{k}{2}} + \frac{1}{(xy)^{\frac{k}{2}}}\right)\left((yz)^{\frac{k}{2}} + \frac{1}{(yz)^{\frac{k}{2}}}\right)\left((zx)^{\frac{k}{2}} + \frac{1}{(zx)^{\frac{k}{2}}}\right).$$

Indeed, this satisfies the given symmetry and conditions, especially the inequality $f(x) < f(y)$ for $1 \le x < y$, due to the strictly increasing nature of $x^k + \frac{1}{x^k}$ for $x > 1$.

### Conclusion

Therefore, the functions that satisfy all given conditions are:
$$f(x) = x^k + \frac{1}{x^k},$$
where $k$ is a positive real number, and should maintain strict monotonicity given the second condition. Thus, the solution is:
$$\boxed{f(x) = x^k + \frac{1}{x^k}}$$ for suitable $k$ such that $f$ is strictly increasing for $x > 1$.