Cauchy-Schwarz inequality in Shankhar's Quantum Mechanics 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
$$ \langle z |= \langle v|- \frac{\langle v|w \rangle}{|w|^2} \langle w|.$$
 A: With regards to your first question, the vector $| z\rangle$ is chosen so it is orthogonal to $| w\rangle$. To see this, note that $\langle a|(|b\rangle+|c\rangle)=\langle a|b\rangle+\langle a|c\rangle$ implies
$$\langle w|z\rangle=\left\langle w\left|v-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle =\langle w|v\rangle+\left\langle w\left|-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle.$$
Let us be careful now about the treatment of the second term, which is related to your second question. We have that $\langle a|(\lambda |b\rangle)\rangle=\lambda\langle a|b\rangle$. Therefore, taking $|a\rangle=|w\rangle=|b\rangle$ and $\lambda=-\frac{\langle w|v\rangle}{|w|^2}$, we see that
$$\left\langle w\left|-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle=-\frac{\langle w|v\rangle}{|w|^2}\left\langle w\left|w\right.\right\rangle$$
Finally, recalling that $|w|^2:=\langle w|w\rangle$, we conclude that
$$\left\langle w\left|-\frac{\langle w|v\rangle}{|w|^2}|w\rangle\right.\right\rangle=-\frac{\langle w|v\rangle}{|w|^2}|w|^2=-\langle w|v\rangle,$$
and
$$\langle w|z\rangle=\langle w|v\rangle-\langle w|v\rangle=0.$$
As a consequence, you get that $|v\rangle$, $|z\rangle$ and $\frac{\langle w|v\rangle}{|w|^2}|w\rangle$ form a right triangle with hypotenuse
$$|v\rangle=|z\rangle +\frac{\langle w|v\rangle}{|w|^2}|w\rangle.$$
Pythagoras' theorem then guarantees that
$$|v|^2=|z|^2+\frac{|\langle w|v\rangle|^2}{|w|^4}|w|^2=|z|^2+\frac{|\langle w|v\rangle|^2}{|w|^2}\geq\frac{|\langle w|v\rangle|^2}{|w|^2}.$$
We conclude that $|v|^2|w|^2\geq|\langle w|v\rangle|^2$. This is the proof given in this Wikipedia article.
If you understood this computation, the answer to your your second question should be clear. Remember that $\langle a|b\rangle=\langle b|a\rangle^*$ and so $\langle (\lambda a)|b\rangle=\lambda^*\langle a|b\rangle$.
Response to edit:
Your first formula for $\langle z|$ is wrong. The second one is the correct one
$$\langle z|=\langle v|-\frac{\langle v|w\rangle}{|w|^2}\langle w|$$
A: If you would take a look at Shankar chapter 1.2 Inner product spaces, he explains that the complex inner product is linear in the ket part and skew-symmetric.
Meaning $\langle v|w\rangle=\langle w|v\rangle^*$ and $\langle v|\alpha w + \beta z\rangle=\alpha \langle v|w\rangle + \beta \langle v | z \rangle $. Now try and work out the expression and you should be fine. 
As for your first question, it is just a definition. I don't know the motivation but I guess it might be a clever one such that the proof works out nicely.
