Strange succession 
An ant starts from the origin of a cartesian plane and makes $n \ge 2$ steps of lengh $d_1, d_2, \cdots, d_n$.  Is there a condition (necessary and sufficient) on $d_i$'s for which the ant can come back to the origin after the $n$ steps? (The ant can move in any direction.)

I think the claim is $d_i \leq \sum\limits_{j ≠ i} d_j$ for all $i$, which of course is necessary, but I cannot prove it is sufficient.

$\def\vec{\overrightarrow}$If the ant is assumed to be smart enough to select the right direction in each step, then the problem can be restated as:

Given $d_1, \cdots, d_n > 0$. Find the necessary and sufficient condition under which there exist vectors $v_1, \cdots, v_n \in \mathbb{R}^2$ such that $|v_k| = d_k$ and $\sum\limits_{k = 1}^n v_k = \vec{0}$.

And your condition is indeed necessary and sufficient.
Necessity: If there exist vectors $v_1, \cdots, v_n \in \mathbb{R}^2$ such that $|v_k| = d_k$ and $\sum\limits_{k = 1}^n v_k = \vec{0}$, then for each $k$,$$
v_k = \sum_{j ≠ k} v_j \Longrightarrow d_k = |v_k| = \left| \sum_{j ≠ k} v_j \right| \leqslant \sum_{j ≠ k} |v_j| = \sum_{j ≠ k} d_j.
$$
Sufficiency: For $n = 2$, $d_1 \leqslant d_2$ and $d_2 \leqslant d_1$ implies $d_1 = d_2$, thus it suffices to take an arbitrary vector $v$ with $|v| = d_1$ and take $v_1 = v$, $v_2 = -v$.
For $n = 3$, since $d_1 \leqslant d_2 + d_3$, $d_2 \leqslant d_1 + d_3$, $d_3 \leqslant d_1 + d_2$, then there exists a triangle $ABC$ (possibly degenerate) with $|AB| = d_1$, $|BC| = d_2$, $|CA| = d_3$. Thus it suffices to take $v_1 = \vec{AB}$, $v_2 = \vec{BC}$, $v_3 = \vec{CA}$.
Now assume that the condition is sufficient for $n - 1$ where $n \geqslant 4$. For $n$, denote $d = \sum\limits_{k = 1}^n d_k$, then the condition becomes $d_k \leqslant \dfrac{1}{2} d$ for each $k$. If there do not exist $k_1 ≠ k_2$ such that $d_{k_1} + d_{k_2} \leqslant \dfrac{1}{2} d$, denoting $d_{n + 1} = d_1$, then$$
2d = \sum_{k = 1}^n (d_k + d_{k + 1}) > n · \frac{1}{2} d \Longrightarrow n < 4,
$$
a contradiction. Therefore, without loss of generality, suppose that $d_{n - 1} + d_n \leqslant \dfrac{1}{2} d$. Using the induction hypothesis on $(d_1, \cdots, d_{n - 2}, d_{n - 1} + d_n)$, there exist $u_1, \cdots, u_{n - 1} \in \mathbb{R}^2$ such that$$
|u_k| = d_k\ (1 \leqslant k \leqslant n - 2),\ |u_{n - 1}| = d_{n - 1} + d_n,\ \sum_{k = 1}^{n - 1} u_k = \vec{0}.
$$
Now take$$
v_k = u_k(1 \leqslant k \leqslant n - 2),\ v_{n - 1} = \frac{d_{n - 1}}{d_{n - 1} + d_n} u_{n - 1},\ v_n = \frac{d_n}{d_{n - 1} + d_n} u_{n - 1},
$$
then $|v_k| = d_k\ (1 \leqslant k \leqslant n)$ and $\sum\limits_{k = 1}^n v_k = \vec{0}$. End of induction.
