I'm trying to understand proof of this inequality. But I have some problems.
So, Shankar starts a proof with definition a new vector $|z \rangle$:
$$ |z \rangle = |v\rangle - \frac{\langle w|v \rangle}{|w|^2} |w \rangle.$$
And here, I would like to know, why we suppose vector in this form? I know that it's projection of one vector on the second one.
My second question is, that we multiply this equation by $\langle z|$. $$ \langle z|z\rangle = \left\langle v - \frac{\langle w|v \rangle}{|w|^2} w \right|\left.v-\frac{\langle w|v \rangle}{|w|^2} |w \right\rangle\\ = \langle v|v \rangle - \frac{\langle w|v \rangle \langle v|w \rangle}{|w|^2} - \frac{\langle w|v \rangle ^*\langle w|v \rangle}{|w|^2} + \frac{\langle w|v \rangle ^*\langle w|v \rangle \langle w|w \rangle}{|w|^4} $$
What I don't understand is a complex conjugate in the last two expression. Why then, we don't conjugate the first and the second one?
EDIT:
So - we're multiplying ket-z times bra-z. So, if the ket-z is: $$ |z \rangle = |v\rangle - \frac{\langle w|v \rangle}{|w|^2} |w \rangle.$$ and bra-z is: $$ \langle z |= \langle v|- \frac{\langle w|v \rangle}{|w|^2} \langle w|.$$
And what concernes me is this term above: $$ \langle w|v \rangle, $$ because if $ | z \rangle = [(\langle z |^*)^T]. $ So shouldn't bra-z be $$ \langle z |= \langle v|- \frac{\langle v|w \rangle}{|w|^2} \langle w|.$$
