Problem 1460

To prove: In fact, in the proof of Example 1, it has already been proven that
$$x+y+z \geq 9xyz$$

Similarly, it is easy to prove that
$$x+y+z \geq \sqrt{3}$$

From Example 1 and the above two inequalities, it is easy to prove this problem. In fact,
on the one hand, we first prove:
$$\frac{27}{4}(x+y)(y+z)(z+x) \geq (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2}$$

By the two-variable mean inequality $2 \sqrt{ab} \leq a+b \left(a, b \in R^{+}\right)$, we get
$$\begin{aligned} & (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \\ = & 2(x+y+z)+2 \sqrt{(x+y)(y+z)}+2 \sqrt{(y+z)(z+x)}+2 \sqrt{(z+x)(x+y)} \\ \leq & 2(x+y+z)+(x+y)+(y+z)+(y+z)+(z+x)+(z+x)+(x+y) \\ = & 6(x+y+z) \end{aligned}$$

Thus, to prove inequality (1), it is only necessary to prove the following inequality.
$$\frac{27}{4}(x+y)(y+z)(z+x) \geq 6(x+y+z)$$

This is clearly a transformation of Example 1, so inequality (**) holds.
On the other hand, we then prove:
$$(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \geq 6 \sqrt{3}$$

By the three-variable mean inequality, Example 1, and $x+y+z \geq \sqrt{3}$, we get
$$\begin{aligned} & (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \\ = & 2(x+y+z)+2[\sqrt{(x+y)(y+z)}+\sqrt{(y+z)(z+x)}+\sqrt{(z+x)(x+y)}] \\ \geq & 2(x+y+z)+2 \cdot 3 \sqrt[3]{\sqrt{(x+y)(y+z)} \cdot \sqrt{(y+z)(z+x)} \cdot \sqrt{(z+x)(x+y)}} \\ = & 2(x+y+z)+6 \sqrt[3]{(x+y)(y+z)(z+x)} \\ \geq & 2(x+y+z)+6 \sqrt[3]{\frac{8}{9}(x+y+z)} \\ \geq & 2 \sqrt{3}+6 \sqrt[3]{\frac{8}{9} \cdot \sqrt{3}} \\ = & 6 \sqrt{3} \end{aligned}$$

Thus, inequality (2) is proven.
In summary, the given inequality holds.