Let $(V_\mathbb R,\langle,\rangle)$ be an inner product space. I’m trying to see for $x,y\in V$ when does $\|x+y\|=\|x\|+\|y\|?$
Let $\|x+y\|=\|x\|+\|y\|$
Squaring both sides, $\langle x+y,x+y\rangle=\langle x,x\rangle+\langle y,y\rangle+2\|x\|.\|y\|\\\ge\langle x,x\rangle+\langle y,y\rangle+2\langle x,y\rangle\\=\langle x,x\rangle+\langle x,y\rangle+\langle x,y\rangle+\langle y,y\rangle\\=\langle x,x+y\rangle+\langle x+y,y\rangle…(*)\\\implies\langle x+y,x+y\rangle-\langle x,x+y\rangle\ge\langle x+y,y\rangle\\\implies0\ge\langle x,0\rangle+\langle y,y\rangle\text{ due to linearity}\\\implies y=0$
Again from $(*),\langle x+y,x+y\rangle-\langle x+y,y\rangle\ge\langle x,x+y\rangle\\\implies 0\ge\langle x,x\rangle+\langle0,y\rangle\\\implies x=0$
Where did I go wrong?
The third line in your formulas can be written (if you use that $\langle x,y\rangle=\langle y,x\rangle$) as $\langle x+y,x+y\rangle$. So you have equality in the inequality in your second line. This implies
$$
\langle x,y\rangle=\|x\|\,\|y\|,
$$
i.e. equality in Cauchy-Schwarz. So $x$ and $y$ are colinear.
Regarding your mistake, the two sides of the inequality in your first “implies” are equal. So your second “implies” is wrong.
Suppose $||x+y|| = ||x|| + ||y||$. Then we have
$$\begin{align*}
\langle x + y, x + y \rangle &= \langle x, x + y \rangle + \langle y, x + y \rangle\\
&= \langle x, x \rangle + 2\langle x, y \rangle + \langle y, y \rangle\\
||x+y||^2&= ||x||^2 + 2\langle x, y \rangle + ||y||^2\\
\end{align*}$$
But from the assumption we have $||x+y||^2 = (||x|| + ||y||)^2 = ||x||^2 + 2||x||\,||y|| + ||y||^2\\$. By the Cauchy–Schwarz inequality, this means x and y are linearly dependent, and specifically $x=cy$ for $c\ge 0$ (notice the absolute value around the inner product in the Cauchy-Schwarz inequality).
I just realized I never addressed your question. The first implication is incorrect, since $\langle x+y,x+y\rangle-\langle x,x+y\rangle=\langle y,x+y\rangle$. In a real inner product space, your “inequality” is an equality, and in a complex vector space it is not safe to assume inequality.