Q:

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

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.

A:

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 $\{ (x_1, x_2, x_3) \in (\mathbb{R}^3)^3 \mid (x_2 - x_1) \times (x_3 - x_1) = 0 \}$.

