Algebra.....or not

Using Cauchy-Schwarz inequality , we have:

$\begin{aligned} (ac+bd)^2 & \le (a^2+b^2)(c^2+d^2) = 2 \times 2 = 4 \\ \implies ac+bd & \le \boxed{2} \end{aligned}$

Equality occurs when $a=b=c=d=1 < \sqrt 2$ .

The sum of squares suggest that we should try constructing right-angled triangles sharing a hypotenuse of length $\sqrt{2}$ .

Notice that the resulting shape is a cyclic quadrilateral ABCD with diameter $\sqrt{2}$ , where AB=a, BC=b, CD=c, DA=d.

Thus, by Ptolemy's Inequality: $ac+bd=AC \times BD=\sqrt{2} \times BD$ . So finding the maximum of BD will give the desired maximum.

Notice that BD is a chord of ABCD's circumcircle, therefore the maximum value of BD is its diameter, which is $\sqrt{2}$ .This is indeed possible when $a=b=c=d=1$

Thus, the maximum value of $ac+bd=\sqrt{2} \times \sqrt{2}=2$ .