Prove infimum of a quantity is fact greater than zero in Euclid space

Question

Assume ${a_i} \in {\mathbb R^3},i = 1,2,3 $

$a_1, a_2, a_3\text{ are not in the same line}$

Prove that:

\begin{array}{l} \inf \left\{ {\sum\limits_{i = 1}^3 {\left\| {x_i - {a_i}} \right\|} \left| {x_i \in \text{a line}} \right.}, i=1,2,3 \right\} > 0 \end{array}

Infimum is taken on every line of $\mathbb R^3$.

I tried but I cannot prove. I am not sure that whether the statement is true. Please help me.

Answer 1

Let $f : (\mathbb{R}^3)^3 \to \mathbb{R}$ be the objective function $f(x_1, x_2, x_3) = \sum_{i=1}^3 \lVert x_i - a_i \rVert$. Then, notice that for each $i,j$, we have that $f(x_1, x_2, x_3) \to \infty$ as $|(x_i)_j| \to \infty$, regardless of the behavior of the other components of vectors. Therefore, there is some $A$ such that for $(x_1, x_2, x_3)$ outside $(B_A(0))^3$ (where $B_A(0)$ is the closed box $[-A, A]^3 \subseteq \mathbb{R}^3$) we have $f(x_1, x_2, x_3) > f(0, 0, 0)$.

Now, $x_1, x_2, x_3$ are colinear if and only if the cross product $(x_2 - x_1) \times (x_3 - x_1) = 0$. This gives a system of three algebraic equations in the coordinates of $x_1, x_2, x_3$, which therefore cuts out a closed subset of $(\mathbb{R}^3)^3$. Now, intersecting this with $(B_A(0))^3$ gives a compact subset of $(\mathbb{R}^3)^3$, so $f$ restricted to this subset achieves a minimum value. By the assumption that $a_1, a_2, a_3$ are not colinear, this minimum value must be greater than 0; and by the previous observations, this minimum value is equal to the overall minimum value of $f$ on $\{ (x_1, x_2, x_3) \in (\mathbb{R}^3)^3 \mid (x_2 - x_1) \times (x_3 - x_1) = 0 \}$.

Answer 2

I think the following is true:

If the $a_i$ are not on a line, then, by the triangle inequality, $\|a_i-a_j\| \lt \|a_i-a_k\|+\|a_j-a_k\| $ for all permutations of ${1, 2, 3}$.

Let $d = \min(\|a_i-a_k\|+\|a_j-a_k\|-\|a_i-a_j\|) $ over all permutations of ${1, 2, 3}$. Then, if the $x_i$ are on a line, $\sum\limits_{i = 1}^3 \left\| {x_i - {a_i}} \right\| \ge d $.

I have the start of a proof, but I don't see how to complete it. Perhaps someone can do that.

Start of proof.

If the $x_i$ are on a line, then $\|x_i-x_j\| = \|x_i-x_k\|+\|x_j-x_k\| $ for some permutation of ${1, 2, 3}$, so that $x_k = tx_i + (1-t)x_j$ for some $0 < t < 1$. Assume this is true for $i=1, j=2, k=3$.

Then

$\begin{array}\\ D &=\sum\limits_{i = 1}^3 \left\| x_i - a_i \right\|\\ &=\| x_1 - a_1 \|+\| {x_2 - a_2} \|+\| {x_3 - a_3} \| \\ &=\| x_1 - a_1 \|+\| x_2 - a_2 \|+\| tx_1 + (1-t)x_2 - a_3 \| \\ &=\| x_1 - a_1 \|+\| x_2 - a_2 \|+\| t(x_1-a_3) + (1-t)(x_2 - a_3) \| \\ \end{array} $

At this point, I don't know what to do. I think that the reverse triangle inequality might be usable, but how is not clear to me.

So, I'll stop here.

Answer 3

The elements of the set of which you are taking the infimum are obviously real numbers greater or equal than zero. So it is enough to show that the infimum is not zero.

Suppose that it is indeed zero. Since that element (as well as any other) is calculated via a sum which terms are $\|x_i - a_i\| \geq 0$ then it follows that every term must be zero, yielding $x_i = a_i$ for $i = 1, 2, 3$. But that is an absurdity since the $x_i$ are in the same line and the $a_i$ are not. Therefore the infimum is not zero.