Result of Cauchy Schwarz inequlaity (homogeneous case) I have shown the Cauchy Schwarz inequality to be $$ \langle v_i(\vec x),v_j(\vec y)\rangle^2 \ \le \ \langle v_i(\vec x),v_i(\vec x)\rangle\langle  v_j(\vec y),v_j(\vec y)\rangle.$$
How would I use this to show $$ \langle v_i(\vec x),v_i(\vec y)\rangle\ \le \ \langle  v_i(\vec x),v_i(\vec x)\rangle$$ for a homogeneous case?
I have considered setting $ _j {_\rightarrow} _i,$ but that doesn't appear to be helpful.

What you want to prove does not hold in general. 
For a counter example, suppose first the vectors have dimension $1$. You want to prove that $(ab)^2\leq a^2 b^2$ implies $(ab)\leq a^2$. However, if $a=1, b=2$, then $2=(ab)>a^2=1$. More generally, for any nonzero vector $x$ and $y=\alpha x$, $\alpha>1$, we have $\langle x,y\rangle = \alpha \langle x,x\rangle >\langle x,x\rangle$.
