Orthonormal basis with specific norm [duplicate]

Assume that $\sum_{i=1}^n a_i v_i = 0$. Then

$$ 0 = \sum_{i=1}^n a_i (v_i - e_i + e_i) \implies \sum_{i=1}^n a_i e_i =\sum_{i=1}^n a_i (e_i - v_i) $$

but then

$$ \sum_{i=1}^n |a_i|^2 = \left| \left| \sum_{i=1}^n a_i e_i \right| \right|^2 = \left| \left| \sum_{i=1}^n a_i(e_i - v_i) \right| \right|^2 \leq \left( \sum_{i=1}^n |a_i| \cdot ||e_i - v_i || \right)^2 \leq \left( \sum_{i=1}^n |a_i|^2 \right) \left( \sum_{i=1}^n ||e_i - v_i||^2 \right) $$

where we used both the triangle inequality for $(V, ||\cdot||)$ and the Cauchy-Schwartz inequality for $\mathbb{R}^n$. This implies that

$$ \left( \sum_{i=1}^n |a_i|^2 \right) \left( \sum_{i=1}^n ||e_i - v_i||^2 - 1 \right) \geq 0 $$

but since $\sum_{i=1}^n||e_i - v_i||^2 < \sum_{i=1}^n \frac{1}{n} = 1$, we must have $\sum_{i=1}^n |a_i|^2 = 0$ as required.