### Solution:

#### Part (1) Proof:

We start by expanding and then subtracting the arithmetic mean of the squares from the square of the arithmetic mean for three elements $a, b, c$:

$$\begin{align*} &{(\frac{a+b+c}{3})}^{2}-\frac{{a}^{2}+{b}^{2}+{c}^{2}}{3}\\ &= \frac{{(a+b+c)}^{2}-3({a}^{2}+{b}^{2}+{c}^{2})}{9}\\ &= \frac{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 3a^2 - 3b^2 - 3c^2}{9}\\ &= \frac{2ab + 2bc + 2ca - 2a^2 - 2b^2 - 2c^2}{9}\\ &= -\frac{{(a-b)}^{2} + {(b-c)}^{2} + {(c-a)}^{2}}{9} \leq 0 \end{align*}$$

Hence, we have proven that ${(\frac{a+b+c}{3})}^{2} \leq \frac{{a}^{2}+{b}^{2}+{c}^{2}}{3}$, and the equality holds if and only if $a=b=c$.

#### Part (2) Finding the minimum value of $t$:

Given the conditions $x > 0$, $y > 0$, $z > 0$, we apply the inequality from part (1) using $\sqrt{x}$, $\sqrt{y}$, and $\sqrt{z}$:

$$\begin{align*} {(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3})}^{2} &\leq \frac{x+y+z}{3}\\ \Rightarrow \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} &\leq \sqrt{\frac{x+y+z}{3}}\\ \Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z} &\leq \sqrt{3(x+y+z)} \end{align*}$$

So, dividing both sides by $\sqrt{x+y+z}$, we get:

$$\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x+y+z}} \leq \sqrt{3}$$

Since $\sqrt{x}+\sqrt{y}+\sqrt{z} \leq t\sqrt{x+y+z}$ must always hold, we find:

$$t \geq \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x+y+z}} \leq \sqrt{3}$$

Thus, the minimum value of $t$ that satisfies this inequality for all positive $x$, $y$, and $z$ is $\boxed{\sqrt{3}}$.