According to a paper by Zhiqin Lu in the Mathematical Gazette (the British publication, not the Boston-area newsletter, if that still exists (or even if it doesn't)) in 2007(?), if $u+v+w=\pi$ and $a,b,c \ge 0$, then $a + b + c \ge 2 \sqrt{bc} \cos u + 2 \sqrt{ca} \cos v + 2 \sqrt{ab} \cos w$. It wasn't hard to show that equality holds iff $a:b:c = \sin^2 u : \sin^2 v : \sin^2 w$, but if I recall correctly, that wasn't in the paper. I would think that there must be some natural geometric interpretation of this proposition. What is it?
The case where equality holds says that if $u+v+w = \pi$, then $\sin^2 u + \sin^2 v + \sin^2 w = 2 \sin u \sin v \cos w + 2 \sin u \cos v \sin w + 2 \cos u \sin v \sin w$. That one has a simple geometric interpretation as a sort of mash-up of the law of sines and the law of cosines. But now suppose we say that if $$\sum_i u_i =\pi$$ then $$ \sum_{i=1}^\infty \sin^2 u_i = \sum_{\text{even }n\ge 2} (-1)^{(n-2)/2}n\sum_{|A|=n} \prod_{i\in A}\sin u_i \prod_{i\not\in A} \cos u_i $$ The case of a sum of three variables that add up to $\pi$ is the special case in which all but three of these are 0. Does it have a geometric meaning?
Later edit: So far we have an answer about the inequality, but not yet about the equality. I will probably comment soon on the former.
