Why must norm induced by an inner product < , > on $C^n$ satisfy the parallelogram law? I know that there is a proof using $||v|| = \sqrt{(< v, v>)} $. But my concern is that why it still holds if ||v|| is defined in other ways? I mean, as far as I know norm is not not necessarily defined as $||v|| = \sqrt{(< v, v>)} $. Hence, I am wondering why proving parallelogram using $||v|| = \sqrt{(< v, v>)} $ means that parallelogram laws apply to all norms?
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The parallelogram law does not apply to all norms. In fact it only applies to norms which are induced by an inner product in the sense that $\| x \| = \sqrt{\langle x,x \rangle}$. The inner product here does not need to be the Euclidean inner product, however.
Many norms do not satisfy the parallelogram law. For example, any $p$-norm for $p \neq 2$ on $\mathbb{C}^n$ does not satisfy the parallelogram law.
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You have a result stating that the space is Euclidian (meaning that the norm has the form $\|X\|^2 = \langle X,X \rangle$ with $\langle . \rangle$ a scalar product) iff the parallelogram identity is true.
To prove this: one way is well known. For the other, assume that the identity is verified.
In these conditions, if $\|X\|^2 = \langle X,X \rangle$ then chack that $$ \langle X,Y \rangle = \frac 12\left[\|X+X^2\| - \|X\| ^2 - \|Y\|^2\right] $$ is a scalar product.