Having a little trouble with this. The Cauchy-Schwatrz inequality for vectors $u$ and $v$ is generally
$$|(u,v)^2|\le (u,u)\cdot(v,v).$$
Then for a two dimensional space its
$$|(u,v)^2|= (\|u\| \|v\|\cos(\theta))^2\le \|u\|^2\|v\|^2.$$
What I did was set two vectors $u$ and $v$ with the values $u=(a,b), v=(x,y).$
Then I used the inequality to plug in the values.
$$||u||=(a^2+b^2),\qquad ||v||=(x^2+y^2),$$
\begin{align*} \cos(\theta)&= (u\cdot v)/(||u||\cdot||v||)\\ &= (ax+by)/((a^2+b^2)\cdot(x^2+y^2)) \end{align*}
$$((a^2+b^2)\cdot (x^2+y^2)\cdot(ax+by)/((a^2+b^2)\cdot(x^2+y^2)))^2\le (a^2+b^2)^2\cdot(x^2+y^2)^2$$
after canceling becomes $$(ax+by)^2\le(a^2+b^2)^2\cdot(x^2+y^2)^2.$$
How could I get rid of the square on the terms on the right side of the inequality?
Edited: Looks like a duplicate but when I searched it did not come up on SE as the inequality was rearranged in the previous question.
Thanks DcMcMor for the editing.
