Motivation for some sort of trig substitution: sum of squares and the fact that everything in the numerator is degree two as well.

and works (in retrospect, this is motivated because stand out like they should be together).



(equality iff ).

So this problem is now equivalent to finding the smallest possible such that

for all .

This is possible iff the discriminant as a quadratic in is nonpositive, or equivalently,





, and it's obviously achievable.

Equality occurs when...

and , so substituting and it becomes the same as the trig solution.

We want to find such that the minimum value of

is .

This means that

(1)

(2)

(3a) and (3b)

(4a) and (4b).

(1)+(2) gives (5).

(3b)+(4a) gives --motivated by elimination of (6).

(3a)+(4b) gives --motivated by elimination of (7).

(6)+(7), and then (5), gives



, and the rest follows. Yay! (Eh, then we can solve the system of equations for the equality stuff).

Eliminate B-3/2

Eliminate C-(rt2-1)/2

Eliminate A-1

Eliminate D-rt2