Prove that 0 is in the convex hull of points chosen from each orthant

If we arbitrarily choose a point from each orthant in $$\mathbb{R}^n$$, that is we choose $$2^n$$ points in total, how do we prove that 0 is in the convex hull of these $$2^n$$ points? It seems obvious, but when I sit down and start thinking about it, I couldn't definitely find a set of $$\lambda_i, i = 1,2,\cdots,2^n$$ such that $$0 \leq \lambda_i \leq 1, \sum_i \lambda_i = 1$$ and $$0 = \sum_i \lambda_i x_i$$, where $$x_i, i = 1,2,\cdots,2^n$$ are any set of points satisfying the condition.

Induct on dimension.

In the $$n=1$$ base case, you have a positive number and a negative number; $$0$$ can be represented as a convex combination of them.

For $$n>1$$: Given your $$2^n$$ points, one from each orthant, divide them into 2 sets of size $$2^{n-1}$$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $$n-1$$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.

Take in pairs the points that differ in sign for the coordinate $$n$$ (and only that one) and form the convex combinations such that that coordinate is cancelled.

Now you have $$2^{n-1}$$ points in each orthants of the diminished space.

The results holds because the convex combination of linear combinations is a convex combination. 