The set of $n>2$ vector is given in the plane, the vector of this set is called long if its length doesn't be less than the length of the vector sum of other $n-1$ vectors.prove if all of the vectors of this set be long, the vector sum of all of them is $\vec{0}$
I think it should be solved using the following structure
consider the smallest vector due to the problem its length shouldn be more than the vector sum of other $n-1$ vector but... how to continue to a solution?
Suppose the sum is $\vec v\neq\vec0$. Place every vector at the origin and rotate all of them so that $\vec v=(v_x,0)$ points along the positive $x$-axis, i.e., $v_x>0$. Because there are $n>2$ vectors, there exists $\vec u=(u_x,u_y)$ such that $u_x<\frac{v_x}2$. Then $$\lVert\vec u\rVert^2=u_x^2+u_y^2<u_x^2+v_x(v_x-2u_x)+u_y^2=(u_x-v_x)^2+u_y^2=\lVert\vec v-\vec u\rVert^2\,,$$ so the length of $\vec u$ is less than than the length of the sum of the other vectors.