Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right? Is $3^x \lt 1 + 2^x + 3^x \lt 3 \cdot 3^x$ right?
This is from my lecture notes which is used to solve: 


But when $x = 0$, $(1 + 2^x + 3^x = 3) \gt (3^0 = 1)$? The thing is how do I choose which what expression should go on the left & right side? 

The inequality is, for $x>0$
$$3^x<1+2^x+3^x<3\cdot 3^x$$
When $x=0$, you obtain $1<1+1+1\le 3\cdot 1$ (I think you have a type in the last part of the post.
But, this is of no concern.  You are trying to find $\lim_{x\rightarrow\infty} (1+2^x+3^x)^{4/x}$. Since the limit is being taken at infinity, you are concerned only with values of $x$ that are large. Large in particular means eventually $x$ is positive, and you can use the inequality above (or, rather, the inequality following it in your post).
The Theorem being used is the following:
Suppose the inequality
$$\tag{1}
f(x)\le g(x)\le h(x)
$$
holds for all $x\ge a$, for some real number $a$.
If $\lim\limits_{x\rightarrow\infty} f(x)$ and  $\lim\limits_{x\rightarrow\infty} h(x)$ exist and are equal with common value $L$, then  $\lim\limits_{x\rightarrow\infty} g(x)$ 
exists and is equal to $L$.
So the Squeeze Theorem is valid whenever the required inequality holds for $x$ sufficiently large in the case where the limit  is taken at infinity.
(If the limit is taken at $b$, then the inequality need onlly holds for $x$ near $b$.)
