Consider a finite dimensional normed vector space $X$. Then $\forall\,x\in X \,\exists\, \{a_1,a_2,...,a_n\}\subset\mathbb R$ such that $x=\sum_{k=1}^na_ke_k$, where $e_k\in\{e_1,e_2,...,e_n\}$ is a basis of $X$.
I want to show that the norm, $ \| x \| = \left ( \sum_{k=1}^{n} | a_k |^2 \right )^{1/2}$, is a genuine norm, although I am having difficulty with the triangle inequality. I see that I should have,
$$ \| x + y \| = \left ( \sum_{k=1}^{n} |a_k + b_k|^2 \right )^{1/2} \leq \left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} + \left ( \sum_{k=1}^{n} |b_k|^2 \right )^{1/2} = \| x \| + \| y \| $$
How, in particular do I make the jump from $\left( \sum_{k=1}^{n} |a_k + b_k|^2 \right )^{1/2}$ to $\left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} + \left ( \sum_{k=1}^{n} |b_k|^2 \right )^{1/2}$? Initially I had proceeded by doing:
$$\left( \sum_{k=1}^{n} |a_k + b_k|^2 \right )^{1/2}\leq\left( \sum_{k=1}^{n} (|a_k| + |b_k|)^2 \right )^{1/2}=\left( \sum_{k=1}^{n} |a_k|^2 +2|a_k||b_k|+ |b_k|^2 \right )^{1/2}$$
But I don't see how this is less than or equal to $\left ( \sum_{k=1}^{n} |a_k|^2 \right )^{1/2} + \left ( \sum_{k=1}^{n} |b_k|^2 \right )^{1/2}$, because of factor of $2|a_k||b_k|$. What am I missing or doing wrong?
Guide:
You want to show that
$$\|x+y\| \le \| x\|+\|y\|$$
squaring both sides, this is equivalent to
$$\|x+y\|^2 \le \|x\|^2 + \|y\|^2+2\|x\|\|y\|$$
Or
$$(x+y)^T(x+y) \le \|x\|^2+\|y\|^2+2\|x\|\|y\|$$
Try to simplify the expression above. You might like to chekck out Cauchy-Schwarz inequality.