Weighted Average Proof Been stuck on this for a while now, seems pretty straightforward but can't seem to prove it.
Given $\mu$ is a weighted average of $\mu_1$ and $\mu_2$ such that $\mu = x_1\mu_1 + x_2\mu_2$ where $x_1$ and $x_2$ are positive real numbers and $x_1 + x_2 = 1$, prove that $\mu$ must lie between $\mu_1$ and $\mu_2$.
And then in the more general case suppose $\mu = x_1\mu_1 + x_2\mu_2 + \cdots + x_n\mu_n$ where again the $x_i$'s are positive and $\sum x_i = 1$, prove that $\mu$ must lie between the smallest and largest of the $\mu_i$'s.

HINT: Without loss of generality assume that $\mu_1\le\mu_2$. Then
$$\mu=x_1\mu_1+x_2\mu_2=x_1\mu_1+x_2\big(\mu_1+(\mu_2-\mu_1)\big)=(x_1+x_2)\mu_1+\ldots$$
Finish this off, and it gives you half of what you want, and a similar idea applied to $\mu_2$ gets you the other hald.
Then try to apply the basic idea to the general case.
