Exercises concerning a linear regression model with parameter $\beta$ I'm trying to solve the following exercises. I am unsure of what I'm doing so I appreciate any help.




My attempt:
(a) We have that
\begin{align*}
\boldsymbol{\hat \beta} &= (\boldsymbol X^T \boldsymbol  X)^{-1} \boldsymbol X^T \boldsymbol Y\\
&= 
\left(\begin{pmatrix}
x_1 & x_2 & \cdots & x_n
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
\vdots\\
x_n
\end{pmatrix}\right)^{-1}
\begin{pmatrix}
x_1 & x_2 & \cdots & x_n
\end{pmatrix}
\begin{pmatrix}
Y_1\\
Y_2\\
\vdots\\
Y_n
\end{pmatrix}\\
&= \Big(\sum_{i = 1}^n x_i^2\Big)^{-1} \Big(\sum_{i = 1}^n x_i Y_i\Big)\\
&= \frac{\sum_{i = 1}^n x_i Y_i}{\sum_{i = 1}^n x_i^2}
\end{align*}
as desired. (Note that $\boldsymbol X^T \boldsymbol  X$ is automatically invertible since it is just a (presumably non-zero) real number.)
(b) This estimator is linear because
\begin{align*}
\tilde{\beta} &= 
\frac{1}{n}
\begin{pmatrix}
\frac{1}{x_1} & \frac{1}{x_2} & \cdots & \frac{1}{x_n}
\end{pmatrix}
\begin{pmatrix}
Y_1\\
Y_2\\
\vdots\\
Y_n
\end{pmatrix},
\end{align*}
that is, it can be written as a linear map.
It is unbiased because $\text E (\tilde{\beta}) = \frac{1}{n} \sum_{i = 1}^n \frac{1}{x_i} \text E(Y_i) = \frac{1}{n} \sum_{i = 1}^n \frac{1}{x_i} \beta x_i = \frac{1}{n} n\beta = \beta$.
Regarding variance, for $\hat \beta$ we have
\begin{align*}
\text{Var}(\hat \beta) &= \text{Var}\Bigg( \frac{\sum_{i = 1}^n x_i Y_i}{\sum_{i = 1}^n x_i^2} \Bigg)\\
&= \frac{1}{\Big( \sum_{i = 1}^n x_i^2 \Big)^2} \text{Var} \Big( \sum_{i = 1}^n x_i Y_i \Big)\\
&= \frac{1}{\Big( \sum_{i = 1}^n x_i^2 \Big)^2} \sum_{i = 1}^n x_i^2 \text{Var}(Y_i)\\
&= \frac{\sigma^2}{\sum_{i = 1}^n x_i^2}.
\end{align*}
For $\tilde \beta$, we have
\begin{align*}
\text{Var}(\tilde \beta) &= \text{Var} \Bigg( \frac{1}{n} \sum_{i = 1}^n \frac{Y_i}{x_i} \Bigg)\\
&= \frac{1}{n^2} \sum_{i = 1}^n \text{Var} \Big( \frac{Y_i}{x_i} \Big)\\
&= \frac{1}{n^2} \sum_{i = 1}^n \frac{1}{x_i^2} \text{Var} (Y_i)\\
&= \frac{\sigma^2}{n^2} \sum_{i = 1}^n \frac{1}{x_i^2}.
\end{align*}
Unless I've made mistakes to this point, I now need to compare the coefficients of $\sigma^2$ in both cases, i.e. compare $\frac{1}{\sum_{i = 1}^n x_i^2}$ with $\frac{1}{n^2} \sum_{i = 1}^n \frac{1}{x_i^2}$, and hopefully see that the latter is at least as large as the former. I am not sure how to do this... maybe I need to try to use an inequality like Cauchy-Schwarz or something.
Edit: Following Glen_b's suggestion, I've used an arithmetic mean/harmonic mean inequality to show
\begin{align*}
\frac{\sum_{i = 1}^n x_i^2}{n} \geq \frac{n}{\sum_{i = 1}^n \frac{1}{x_i^2}}\\
\Rightarrow \frac{1}{n^2} \sum_{i = 1}^n \frac{1}{x_i^2} \geq \frac{1}{\sum_{i = 1}^n x_i^2}
\end{align*}
as desired.
(c) I am not sure what to do here. Do I have to be able to say something like ``$\text{Var}(\epsilon_i) = $ [something], and $\text{Cov}(\epsilon_i, \epsilon_j) = $ [something]?
Thank you.

I assume the $x_i$'s are fixed constants.
Consider uncorrelated but heteroscedastic errors , i.e.
$$\operatorname{Cov}(\varepsilon_i,\varepsilon_j)=
\begin{cases}\sigma_i^2 &,\text{ if }i=j
\\ 0 &, \text{ if }i\ne j
\end{cases}$$
Assuming of course $\sigma_i>0$ for every $i$, rewrite the  model as
$$\frac{Y_i}{\sigma_i}=\beta \frac{x_i}{\sigma_i}+\frac{\varepsilon_i}{\sigma_i} \quad\text{ or }, \quad Y_i^*=\beta x_i^*+\varepsilon_i^*$$
This model with $Y_i^*$ as response and $x_i^*$ as covariate satisfies the Gauss-Markov assumptions, so that OLS can be performed on this transformed model. This leads to the weighted least squares estimator of $\beta$:
$$\tilde \beta=\frac{\sum_{i=1}^n x_i^* Y_i^*}{\sum_{i=1}^n (x_i^*)^2}=\frac{\sum\limits_{i=1}^n \frac{x_i Y_i}{\sigma_i^2}}{\sum\limits_{i=1}^n \frac{x_i^2}{\sigma_i^2}}$$
If you choose $\sigma_i^2 \propto x_i^2$, i.e. $\sigma_i^2=\sigma^2 x_i^2$ for some positive $\sigma$, you get the estimator in your post:
$$\tilde \beta=\frac1n\sum_{i=1}^n \frac{Y_i}{x_i}$$
