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$$\left\|\widehat{e_1}+\widehat{e_2}+\cdots+\widehat{e_n}\right\|<\left\|\widehat{e_1}\right\|+\left\|\widehat{e_2}\right\|+\cdots+\left\|\widehat{e_n}\right\|.$$