I need some help on this question. I just have no idea on how to get started on this problem.
Here is the problem:
For two and three dimensional vectors, the fundamental property of dot products
$ A \cdot B = |A||B| \cos{\theta}$ implies that $A \cdot B \leq |A||B|$ (1.1)
Show that $|A- \gamma B|^2$ implies (1.1) where $\gamma = \frac{A \cdot B}{B \cdot B}$
I just do not know how to begin with this problem
If $|B|=0$, then $|A||B|\geq|A\cdot B|$ holds trivially. Else, consider: $$ 0\leq |A-\gamma B|^2=|A|^2-2\gamma A\cdot B+\gamma^2|B|^2. $$ Now, if you use $\gamma=\frac{A\cdot B}{B\cdot B}$, then the above is translated to $$ 0\leq |A|^2-2\frac{(A\cdot B)^2}{|B|^2}+\frac{(A\cdot B)^2}{|B|^2}=|A|^2-\frac{(A\cdot B)^2}{|B|^2}\implies |A|^2\geq \frac{(A\cdot B)^2}{|B|^2} $$ which rearranges to give $|A|^2|B|^2\geq(A\cdot B)^2$, or, equivalently, $|A||B|\geq|A\cdot B|$.