Have to minimize $f(x, y, z) = (x-1)^2 + (\frac{y}{x} -1)^2+(\frac{z}{y} -1)^2+(\frac{4}{z} -1)^2$ on the set $\{(x, y, z) : 1 \le x \le y \le z \le 4\}$.
Lagrange multiplier needs to be used but I'm quite lost on the particular set mentioned. I would appreciate some help.
To apply Lagrange multipliers, you
The Lagrange multipliers will give you something like
$$u_k - 1 = \frac{\lambda}{u_k} \prod_{j=1}^4 u_j\quad\text{ for some constant }\lambda$$ which is equivalent to $$u_1(u_1 - 1) = u_2(u_2 - 1) = u_3(u_3 - 1) = u_4(u_4 - 1)$$ This in turn implie $u_2, u_3, u_4 \in (u_1,1-u_1)$. Since $u_1 = x$ is supposed to be $\in [1,4]$, $1 - u_1$ will be non-positive and this rules out the possibility that $u_2, u_3, u_4 = 1 - u_1$. As a result, the methods of Lagrange multipliers suggests the minimum may be located at a configuration where $u_1 = u_2 = u_3 = u_4$. Since $\prod\limits_{k=1}^4 u_k = 4$, the common value of $u_k$ should be $\sqrt{2}$.
In general, when you apply Lagrange multipliers to minimize something, there are usually many loose ends in the logic. To verify above ansatz "all $u_k = \sqrt{2}$" indeed give us the true minimum, we apply AM $\ge$ GM to the 4 numbers $u_k$, we obtain
$$\sum_{k=1}^4 u_k \ge 4 \left(\prod_{k=1}^4 u_k\right)^{\frac14} = 4\sqrt[4]{4} = 4\sqrt{2}$$
In terms of $u_k$, we have
$$\begin{align} f(x,y,z) = & \sum_{k=1}^4 (u_k - 1)^2\\ = & \sum_{k=1}^4 \left((\sqrt{2}-1)^2 + (u_k - \sqrt{2})^2 + 2(\sqrt{2}-1)(u_k-\sqrt{2})\right)\\ = & 4(\sqrt{2}-1)^2 + \sum_{k=1}^4 (u_k-\sqrt{2})^2 + 2(\sqrt{2}-1)\left(\sum_{k=1}^4 u_k - 4\sqrt{2}\right)\\ \ge & 4(\sqrt{2}-1)^2 \end{align} $$ This means $f(x,y,z)$ is bounded from below by $4(\sqrt{2}-1)^2$. Since this lower bound is attained at $(x,y,z) = (\sqrt{2},2,2\sqrt{2})$ which belongs to the given set. $f(x,y,z)$ does attain its minimum at a configuration where $u_1 = u_2 = u_3 = u_4 = \sqrt{2}$ with value equals to $4(\sqrt{2}-1)^2$.