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).

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 $b=d$.
