Let n be a non zero natural number and n vectors in plane $\vec{v_{1}},\vec{v_{2}},...,\vec{v_{n-1}},\vec{v_{n}}$.Prove that there exist $a_{1},a_{2},a_{3},...,a_{n}\in\left \{ 1,-1 \right \}$ for that: $$\left | a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+...+a_{n}\vec{v_{n} }\right |^{2}\geqslant\ \left | \vec{v_{1}} \right |^{2}+\left | \vec{v_{2}} \right |^{2}+...+\left | \vec{v_{n}} \right | ^{2}$$
My solution: By induction
Suppose there is $ n = 1 $. It's trivial
So $n'=n+1$
$1 \leq k \leq n$
$1 \leq k' \leq n'$
$$|\sum ak' \sqrt{k'}| \geq |\sum ak \sqrt{k}|+ |an' \sqrt{n}|$$ $$|\sum ak' \sqrt{k'}|^2 \geq |\sum ak \sqrt{k}|^2+2||\sum ak \sqrt{k}|| an' \sqrt{n}|+|an' \sqrt{n}'|^2 \geq |\sum ak \sqrt{k}|^2+|an' \sqrt{n}|^2 \geq \sum|ak' \sqrt{k'}|^2$$
Correct?
Consider just two vectors $\vec u$ and $\vec v$, and a constant $c \in \{-1, 1\}$ to be determined.
$$\begin{align*} \left|\vec u +c\vec v\right|^2 &= \left(\vec u +c\vec v\right)\cdot \left(\vec u +c\vec v\right)\\ &= \vec u\cdot \vec u + 2c\left(\vec u\cdot \vec v\right) + c^2\left(\vec v \cdot \vec v\right)\\ &= \left|\vec u \right|^2 + \left|\vec v \right|^2 + 2c\left(\vec u\cdot \vec v\right) \end{align*}$$
By choosing $c$ to match the sign of $\vec u\cdot \vec v$, or if $\vec u\cdot \vec v = 0$ then $c=1$,
$$\begin{align*} c\left(\vec u\cdot \vec v\right) &\ge 0\\ \left|\vec u +c\vec v\right|^2 &\ge \left|\vec u \right|^2 + \left|\vec v \right|^2 \end{align*}$$
i.e. for any two vectors $\vec u, \vec v$, there exists a $c\in\{-1, 1\}$ such that $\left|\vec u +c\vec v\right|^2 \ge \left|\vec u \right|^2 + \left|\vec v \right|^2$.
Back to induction step. Assume that for some $n = k$, there exists $a_1, a_2, \ldots, a_k \in \{-1, 1\}$ s.t.
$$\left|\sum_{i=1}^k a_i\vec {v_i}\right|^2 \ge \sum_{i=1}^k \left|\vec{v_i}\right|^2$$
For $n = k+1$, using the result for two vectors, choose an $a_{k+1} \in \{-1, 1\}$ s.t.
$$\left|\left(\sum_{i=1}^k a_i\vec {v_i}\right) + a_{k+1} \vec {v_{k+1}}\right|^2 \ge \left|\sum_{i=1}^k a_i\vec {v_i}\right|^2 + \left|\vec{v_{k+1}}\right|^2$$
Then using this $a_{k+1}$ and the induction hypothesis,
$$\begin{align*} \left|\sum_{i=1}^{k+1} a_i\vec {v_i}\right|^2 &= \left|\left(\sum_{i=1}^k a_i\vec {v_i}\right) + a_{k+1} \vec {v_{k+1}}\right|^2\\ &\ge \left|\sum_{i=1}^k a_i\vec {v_i}\right|^2 + \left|\vec{v_{k+1}}\right|^2\\ &\ge \sum_{i=1}^k \left|\vec{v_i}\right|^2 + \left|\vec{v_{k+1}}\right|^2\\ &= \sum_{i=1}^{k+1} \left|\vec{v_i}\right|^2 \end{align*}$$
Together with the base case when $n=1$, this finishes the proof.