Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$ Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$
I seem to struggle with this simple proof. All I managed to find is that ac+bd=-4 (which might not even be correct).
 A: You can use Cauchy Schwarz inequality:
$$(ac+bd)^2 \leq (a^2+b^2)\cdot(c^2+d^2)$$
$$(ac+bd) \leq 1 $$
A: $(ac+bd)^2 - (a^2+b^2)(c^2+d^2) = -(ad-bc)^2 \leq 0 \to (ac+bd)^2 \leq 1 \to ...$
A: This is asking you to show that the dot product of two $2$-dimensional unit vectors is at most one, and follows immediately if you know about a certain formula related to the dot product:
$$
ac + bd = \langle a, b \rangle \cdot \langle c, d \rangle = \|\langle a, b \rangle\| \|\langle c, d \rangle\| \cos\theta = \sqrt{a^2 + b^2} \sqrt{c^2 + d^2}\cos\theta = \cos\theta \leq 1
$$
(where $\theta$ is the angle between the two vectors). This is basically just the Cauchy-Schwarz inequality:
$$
ac + bd \leq \sqrt{a^2 + b^2} \sqrt{c^2 + d^2} = 1
$$
A: hint: try to compute $(a-c)^2+(b-d)^2$, and evaluate some inequalities :)
A: Alternatively write those numbers in terms of sines and cosines and use the sum formula.
To be more precise, write $a=\sin x,b=\cos x,d=\sin y,c=\cos y$ and note $ac+bd=\sin(x+y)$, in fact also $\cos(x+y)=bc-ad$ so we obtain $$a^2+b^2=1,c^2+d^2=1\implies (ac+bd)^2+(ad-bc)^2=1$$
This can also be seen if we consider complex multiplication. As user Aryabhata noted, we're showing $|z|=|w|=1\implies |zw|=1$. We can show more generally that the complex norm is multiplicative, of course. 
A: 1st Method
$\begin{align}\left(a^2+b^2\right)\left(c^2+d^2\right)=1& \implies (ac+bd)^2+(ad-bc)^2=1\\&\implies (ac+bd)^2\le1\end{align}$
2nd Method
$\begin{align}\left(a^2+b^2\right)+\left(c^2+d^2\right)=2& \implies \left(a^2+c^2\right)+\left(b^2+d^2\right)=2\\&\implies 2(ac+bd)\le 2\qquad  \text{(by A.M.- G.M. Inequality)}\\&\implies (ac+bd)\le 1\end{align}$
A: Here's another cool geometric approach if $a,b,c,d$ are positive. Consider the following diagram where $BD=1$ and is the diameter of the circle.

By Pythagoras, we have $a^2+b^2=1$ and $c^2+d^2=1$. However, by Ptolemy's Theorem, we also have $ac+bd=BD\cdot AC=AC$. Since $AC$ is any chord in a circle with diameter $1$, we have $AC\leq 1$. Thus, $ac+bd=AC\leq 1$ as desired. We can also find that equality occurs when $AC$ is also a diameter, which means $a=c$ and $b=d$.
