How to find the tightest bounds of sequences using squeeze theorem?

I am trying to determine how to find the tightest upper and lower bounds when using the squeeze theorem. If there is a general technique to determining them I would appreciate the insight. Specifically I have this problem:

$$\lim_{n\to \infty} \sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n}$$

I know that:

$$\sqrt[n]{\left(\frac23\right)^n}\le \sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n} \le \sqrt[n]{6\left(\frac12\right)^n}$$

What steps were taken to find the upper bound?

You want to bound your expression above and below with other expressions that are simpler and converge to the same limit. Since we have a $n$-th root and terms raised to the $n$-th power, it seems reasonable to try and use the fact that $\sqrt[n]{x^n}=x$.
For the lower bound simply note that $\sqrt[n]{x}$ is increasing and $a^n\ge 0$ for $a>0$. Then, $$\sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n} \ge 0+\sqrt[n]{\left(\frac23\right)^n+0}=\frac23$$ For the upper bound note that $\left(\frac12\right)^n\le \left(\frac23\right)^n$ for $n>0$. Therefore, $$\sqrt[n]{2\left(\frac12\right)^n+\left(\frac23\right)^n+3\left(\frac12\right)^n} \le \sqrt[n]{2\left(\frac23\right)^n+\left(\frac23\right)^n+3\left(\frac23\right)^n}=\frac23\sqrt[n]{6}$$ The squeeze theorem will now do the trick (note that $\sqrt[n]a\to 1$)