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)&amp;= (u\cdot v)/(||u||\cdot||v||)\\ &amp;= (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.
Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.  So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$  and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$  So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure how to show that
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