Show $\exists u,v \in X$ such that $\|u+v\|^2 \neq \|u\|^2+\|v\|^2$

Question

Let $(X, \langle \cdot , \cdot \rangle )$ be an inner product space. Show that if $X \neq {0}$ then there exist $u,v∈X$ such that $\|u+v\|^2 \neq \|u\|^2+\|v\|^2$.

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Can we just say that choose $u$ and $v$ such that they are not orthogonal to each other? Then it would work I think but I can't think of specific examples because it is an arbitrary norm.

Answer 1

If $X\neq 0$, it contains a nonzero vector $v$. It also contains $-v$. What is $\Vert v+(-v)\Vert^2$? What is $\Vert v\Vert^2+\Vert-v\Vert^2?$

Answer 2

Suppose for all $\;u,v\in X\;$ we have

$$\left\|u\right\|^2+\left\|v\right\|^2=\left\|u+v\right\|^2:=\left\|u\right\|^2+\left\|v\right\|^2+2\,\text{Re}\,\langle u,v\rangle\iff\text{Re}\,\langle u,v\rangle=0\;\;\;\forall\,u,v\in X$$

But this is absurd since there exists $\;0\neq x\in X\implies \langle x,x\rangle=\text{Re}\,\langle x,x\rangle=\left\|x\right\|^2\neq0\;$