I have reason to believe the text has a typo; maybe someone can correct me on this point. Because, to my mind, the definition of the $\hat{c}_i$'s would apply to the sum $\sum |a_k b_k c_k^2|$. I suspect it should read
$$\hat{c}_i=\frac{c_i}{\sqrt{c_1^2+c_2^2+\cdots+c_n^2}}.$$
If my hunch is correct, we would argue as follows (using the $\hat{c}_i$'s defined right above):
$$\left|\sum_{k=1}^n a_kb_k\hat{c}_k \right|
\color{Red}\le \sum_{k=1}^n |a_kb_k \hat{c}_k|
\color{Green}\le \sum_{k=1}^n |a_kb_k|
\color{Blue}=\left|\sum_{k=1}^n |a_k|\cdot|b_k|\right|
\color{Purple}\le \left(\sum_{k=1}^n |a_k|^2\right)^{1/2}\left(\sum_{k=1}^n |b_k|^2\right)^{1/2} $$
Above:
- $\color{Red}\le$: Follows from triangle inequality
- $\color{Green}\le$: Follows from $|\hat{c}_k|\le1$, $k=1,\cdots,n$.
- $\color{Blue}=$: Follows because $x=|x|$ for $x\ge0$.
- $\color{Purple}\le$: Cauchy-Schwarz applied to $|a_k|,|b_k|$.
Now take the far left and far right side of this, square, and multiply by $c_1^2+c_2^2\cdots+c_n^2$ (apply to $\hat{c}_i$).
It is not entirely clear what you mean by 'following Tao's argument'.
Suppose $|\langle v,w\rangle| = \|v\| \|w\|$.
Let $w=\alpha v +h$, with $h \bot v$. Then $\|w\|^2 = |\alpha|^2 \|v\|^2 + \|h\|^2$ and $|\langle v,w\rangle| = |\alpha| \|v\|^2$.
Then the first squared line gives:
$|\alpha|^2 \|v\|^4 = \|v\|^2 (|\alpha|^2 \|v\|^2 + \|h\|^2)$.
Hence either $v=0$ and we have $v = 0 \cdot w$, or $h=0$ in which case $w = \alpha v$.
Alternative approach:
If $w=0$ or $v=0$ the result is true so suppose both are not zero.
Suppose $|\langle v,w\rangle| = \|v\| \|w\|$. Replacing $v$ by $\theta v$, with $|\theta| = 1$ does not change the formula, so
we can assume that
$\langle v,w\rangle = \|v\| \|w\|$.
Note that replacing $(v,w)$ by $(tv, {1 \over t} w)$, with $t>0$ does not change the formula.
Now note that $(t\|v\|-{1 \over t}\|w\|)^2 = \|tv-{1 \over t}w\|^2$.
Now choose $t=\sqrt{\|w\| \over \|v\|}$ to get
$w = {\|w\| \over \|v\|} v$.
Best Answer
$(x+y+z)^2 = \left(x\sqrt{\frac{y+z}{y+z}} +y\sqrt{\frac{x+z}{x+z}}+z\sqrt{\frac{x+y}{x+y}}\right)^2\leq\left(\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y}\right)(y+z+x+z+x+y)$. You will get it. I didn't read that book, but I believe Srivatsan is right.