India — Team Selection Test, 2009

We prove the result for a more general class of vectors having magnitude not exceeding $1$. We use induction on $n$. If $n = 2$, parallelogram law gives
$$|v_1 + v_2|^2 + |v_1 - v_2|^2 = 2(|v_1|^2 + |v_2|^2) \le 4.$$
Hence either $|v_1 + v_2| \le \sqrt{2}$ or $|v_1 - v_2| \le \sqrt{2}$, giving the result. Suppose $n \ge 3$ and the result is true for $n-1$ vectors. Consider $n$ vectors $v_1, v_2, \dots, v_n$ such that $|v_j| \le 1$, for $1 \le j \le n$. Among the six vectors $\pm v_1, \pm v_2, \pm v_3$, there are two vectors, the angle $\theta$ between which is $\le 60^\circ$. If we denote these two by $u_1$ and $u_2$, we have
$$\begin{aligned} |u_1 - u_2|^2 &= |u_1|^2 + |u_2|^2 - 2|u_1||u_2| \cos \theta \\ &\le 2 - 2 \cos \theta \\ &\le 2 - 2 \cos 60^\circ = 1. \end{aligned}$$
Thus $|u_1 - u_2| \le 1$. After renaming if necessary, we can take $u_1 - u_2 = \lambda_1' v_1 + \lambda_2' v_2$. Now consider $n-1$ vectors $u_1 - u_2, v_3, \dots, v_n$. Induction hypothesis applies to this set of vectors. We therefore get
$$|\lambda'(\lambda_1' v_1 + \lambda_2' v_2) + \lambda_3 v_3 + \dots + \lambda_n v_n| \le \sqrt{2}.$$
Taking $\lambda_j = \lambda'\lambda'_j$, for $j = 1, 2$, we get
$$|\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 + \dots + \lambda_n v_n| \le \sqrt{2}.$$