Prove that if $\vec v$ and $\vec w$ are orthogonal then $\|\vec v+\vec w\|=\|\vec v\|+\|\vec w\|$

Question

So if $\vec V$ is a real vector space equipped with an inner product $\langle\cdot,\cdot\rangle$, then using the polar identity, it is not hard to show that the equality $$ \|\vec v+\vec w\|=\|\vec v\|+\|\vec w\| $$ holds if and only if $$ \langle\vec v,\vec w\rangle=\|\vec v\|\cdot\|\vec w\| $$ and by the Cauchy-Schwarz theorem the last equality holds if and only the vectors $\vec v$ and $\vec w$ are linearly dependent. So if $\vec v$ and $\vec w$ are orthogonal we conclude that the first equality does not holds, that is $$ \|\vec v+\vec w\|<\|\vec v\|+\|\vec w\| $$ by the triangular inequality. However this it seems to me strange because the Pythagoras' theorem must hold for any $n$ orthogonal vectors: indeed if the vectors $\hat e_i$ for $i=1,...,n$ are the vectors of the canonical basis then $$ \|\hat e_1+...+\hat e_n\|=\sqrt n=\underbrace{1+...+1}_{\text{n times}}=\|\hat e_1\|+...+\|\hat e_n\| $$ and clearly this holds for any orthonormal basis. So why my first argument is incorrect? and how to prove the statement? Could someone help me, please?

Answer 1

It turns out that$$\sqrt n<\overbrace{1+1+\cdots+1}^{n\text{ times}}=n$$and that therefore$$\left\|\widehat{e_1}+\widehat{e_2}+\cdots+\widehat{e_n}\right\|<\left\|\widehat{e_1}\right\|+\left\|\widehat{e_2}\right\|+\cdots+\left\|\widehat{e_n}\right\|.$$

Answer 2

There's something wrong with your equivalent statements. If it were true: say $v=1$ and $w=-1$ which are linearly dependent.

Equivalently (?), $-1=\langle v,w\rangle =||v||\cdot ||w||=1$?

Equivalently (?), $0=||v+w||=||v||+||w||=2$?