"Show that T is an unbiased estimator" follow-up

Question

Yesterday I posted this question Show that $T$ is an unbiased estimator and with the help of others, I got to understand why $T=\frac{n+1}{2n+1}X_{(n)}$ is an unbiased estimator of $\theta$, with regard to the PDF $f(x)$ in the question. Following the same reasoning, I proved the second part of the exercise, that $U=\frac{n+1}{5n+4}(2X_{(n)}+X_{(1)})$ is also an unbiased estimator of $\theta$. So, we have $E[T]=E[U]=\theta$.

Now, onto the thrid part of the exercise, where I need a bit of help. I am asked to show that $Var(T)\geqslant Var(U)$. I choosed the straight-forward path (using $Var(X)=E[X^2]-(E[X])^2$) ; $$Var(T)\geqslant Var(U) \iff E[T^2]-(E[T])^2 \geqslant E[U^2]-(E[U])^2 \iff E[T^2] \geqslant E[U^2]$$ since $(E[T])^2 = (E[U])^2 = \theta$. Moving on, $$E[\left(\frac{n+1}{2n+1}X_{(n)}\right)^2] \geqslant E[\left(\frac{n+1}{5n+4}(2X_{(n)}+X_{(1)})\right)^2]$$ $$\iff \left(\frac{n+1}{2n+1}\right)^2 E[(X_{(n)})^2] \geqslant \left(\frac{n+1}{5n+4}\right)^2 E[(2X_{(n)}+X_{(1)})^2$$ $$\iff \left(\frac{1}{2n+1}\right)^2 E[(X_{(n)})^2] \geqslant \left(\frac{1}{5n+4}\right)^2 (E \left[4(X_{(n)})^2+ 4X_{(1)}X_{(n)}+ (X_{(1)})^2 \right]$$ $$\iff \left(\frac{5n+4}{2n+1}\right)^2 E[(X_{(n)})^2] \geqslant 4E(X_{(n)})^2 ]+ 4E[X_{(1)}X_{(n)} ]+ E[(X_{(1)})^2 ]$$ $$\iff \left( \left(\frac{5n+4}{2n+1}\right)^2 -4\right)E(X_{(n)})^2 ] \geqslant 4E[X_{(1)}X_{(n)} ]+ E[(X_{(1)})^2 ]$$

Now, we know that if $X\geqslant Y$, then $E[X] \geqslant E[Y]$. Since $X_{(n)}=max{X_i}, i=1,2,....,n$, then it is obvious that $E(X_{(n)})^2 ] \geqslant E[X_{(1)}X_{(n)} ]$ and that $E(X_{(n)})^2 ] \geqslant E[(X_{(1)})^2 ]$ (all of the variables are positive, if that matters). So, $$4E[X_{(1)}X_{(n)} ]+E[(X_{(1)})^2 ] \leqslant 5E[(X_{(n)})^2 ]$$ Therefore, if I manage to show that $$\left( \left(\frac{5n+4}{2n+1}\right)^2 -4\right)E[(X_{(n)})^2 \geqslant 5E(X_{(n)})^2 ]$$ the proof will be over. $$\left( \left(\frac{5n+4}{2n+1}\right)^2 -4\right)E[(X_{(n)})^2 \geqslant 5E(X_{(n)})^2 ]$$ $$\iff \left(\frac{5n+4}{2n+1}\right)^2 -4 \geqslant 5$$ $$\iff \left(\frac{5n+4}{2n+1}\right)^2 \geqslant 9=3^2$$ $$\iff \frac{5n+4}{2n+1} \geqslant 3$$ (the negative solutions are not acceptable, since $n$ is the number of the sample we get, hence positive) $$\iff 5n+4 \geqslant 6n+3 \iff n \leqslant 1$$ Now, clearly that cannot be, in fact I should get exactly the opposite inequality $n \geqslant 1$. So, two things may be happening; either $a)$ The exercise has a typo, and $Var(T)\leqslant Var(U)$ and not "$\geqslant$", or $b$) I have made a mistake, either in my reasoning or in the calculations.

Once again, any help would be really helpful, to confirm that the exercise indeed has a typo or to find out what the mistake is. Thanks in advance (and sorry for the long, long text)!

Answer 1

Your approach is unsuccessful because of "if I manage to show that". You are replacing the desired inequality with a stronger inequality that is harder to satisfy, i.e., the stronger inequality fails to be true for all $n$. It is akin to arguing

I want to prove $n^2+1\ge n^2$ for all $n$. I know that $n^2+1\ge 1$ for all $n$, so if I can show $1\ge n^2$, then I am done.

Instead, it is straightforward to prove the desired inequality as stated, by explicitly substituting $E(X_{(1)})^2$, $E(X_{(n)})^2$, and $E(X_{(1)}X_{(n)})$. You can compute the first two from the (marginal) densities of $X_{(1)}$ and $X_{(n)}$ respectively, and the third via the joint density of the pair $(X_{(1)}, X_{(n)})$: $$f_{X_{(1)}, X_{(n)}}(x,y) = n(n-1)(y-x)^{n-2}, \quad 0 \le x \le y \le 1.$$ If needed, you can deduce the marginal densities from the joint density.