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.
Best Answer
In an ancient MathLinks topic (post #6; but see below for a copy) I have given a proof of the inequality by reducing it to $\left(\sqrt{a}\vec{p}+\sqrt{b}\vec{q}+\sqrt{c}\vec{r}\right)^2\geq 0$, where multiplication means scalar product of vectors and $\vec{p}$, $\vec{q}$, $\vec{r}$ are unit length vectors chosen in such a way that the angles between them are $\pi-u$, $\pi-v$, $\pi-w$, respectively. This rewrites geometrically as follows: Pick a point $P$ in the plane, and take three points $A$, $B$, $C$ such that $PA=\sqrt{a}$, $PB=\sqrt{b}$, $PC=\sqrt{c}$, $\measuredangle BPC=\pi-u$, $\measuredangle CPA=\pi-v$ and $\measuredangle APB=\pi-w$. Then, the difference between the left hand side and the right hand side of your inequality is $9$ times the square of the distance between the point $P$ and the centroid of triangle $ABC$. Equality thus holds if and only if $P$ is the centroid of triangle $ABC$; this is equivalent to the assertion that the triangles $BPC$, $CPA$, $APB$ have equal areas; this, in turn, is equivalent to the assertion that $\sqrt{a}:\sqrt{b}:\sqrt{c}=\sin u:\sin v:\sin w$ (because the area of triangle $BPC$ is $\frac{1}{2}\cdot PB\cdot PC\cdot \sin\measuredangle BPC=\frac{1}{2}\sqrt{b}\sqrt{c}\sin u$ etc.).
For better searchability, let me copy my MathLinks posts over here (finding some old post on MathLinks is almost impossible as for now). Note that I do not claim originality for the theorems.
Theorem 1. Let $x$, $y$, $z$ be three real numbers and $A$, $B$, $C$ three real angles such that $A + B + C = 180^{\circ}$. Then,
$x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$.
Proof of Theorem 1. We will denote by $\measuredangle\left(\overrightarrow{p};\;\overrightarrow{q}\right)$ the directed angle between two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ (note that this is a directed angle modulo $360^{\circ}$).
For any two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$, we are going to denote by $\overrightarrow{p}\cdot\overrightarrow{q}$ the scalar product of the vectors $\overrightarrow{p}$ and $\overrightarrow{q}$.
For any vector $\overrightarrow{p}$, we are going to denote by $\overrightarrow{p}^2$ the scalar product $\overrightarrow{p}\cdot\overrightarrow{p}$. Every vector $\overrightarrow{p}$ satisfies $\overrightarrow{p}^2=\left|\left|\overrightarrow{p}\right|\right|^2\geq 0$.
Let $\overrightarrow{a}$ be a vector of unit length. Let $\overrightarrow{b}$ be a vector of unit length such that $\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)=180^{\circ}-C$. Let $\overrightarrow{c}$ be a vector of unit length such that $\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$. Then,
$\measuredangle\left(\overrightarrow{c};\;\overrightarrow{a}\right)=360^{\circ}-\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)-\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)$ $=360^{\circ}-\left(180^{\circ}-C\right)-\left(180^{\circ}-A\right)=C+A=180^{\circ}-B$
(since $A + B + C = 180^\circ$).
Now, all the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ have unit length: $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$. Thus, $\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$ yields
$\overrightarrow{b}\cdot\overrightarrow{c}=\left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|\cdot\cos\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=1\cdot 1\cdot\cos\left(180^{\circ}-A\right)=\cos\left(180^{\circ}-A\right)$ $=-\cos A$.
Similarly, we obtain $\overrightarrow{c}\cdot\overrightarrow{a}=-\cos B$ and $\overrightarrow{a}\cdot\overrightarrow{b}=-\cos C$. Thus,
$\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2$
$=\left(x\cdot\overrightarrow{a}\right)^2+\left(y\cdot\overrightarrow{b}\right)^2+\left(z\cdot\overrightarrow{c}\right)^2$
${}+2\cdot y\cdot\overrightarrow{b}\cdot z\cdot\overrightarrow{c}+2\cdot z\cdot\overrightarrow{c}\cdot x\cdot\overrightarrow{a}+2\cdot x\cdot\overrightarrow{a}\cdot y\cdot\overrightarrow{b}$
$=x^2\underbrace{\cdot\left|\overrightarrow{a}\right|^2}_{=1^2}+y^2\cdot\underbrace{\left|\overrightarrow{b}\right|^2}_{=1^2}+z^2\cdot\underbrace{\left|\overrightarrow{c}\right|^2}_{=1^2}+2yz\cdot\underbrace{\overrightarrow{b}\cdot\overrightarrow{c}}_{=-\cos A}+2zx\cdot\underbrace{\overrightarrow{c}\cdot\overrightarrow{a}}_{=-\cos B}+2xy\cdot\underbrace{\overrightarrow{a}\cdot\overrightarrow{b}}_{=-\cos C}$
$=x^2\cdot 1^2+y^2\cdot 1^2+z^2\cdot 1^2+2yz\cdot\left(-\cos A\right)+2zx\cdot\left(-\cos B\right)+2xy\cdot\left(-\cos C\right)$
$=x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C$.
Since we, obviously, have $\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2\geq 0$, we thus get $x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C\geq 0$, so that $x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$, and Theorem 1 is proven.
Other proofs of Theorem 1 can be found at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5243 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42509 .
Theorem 1 is equivalent to the following, also quite useful (for olympiad mathematics and magazine problem sections, that is, although I would not be surprised to see more applications) inequality:
Theorem 2. Let $x$, $y$, $z$ be three real numbers and $A$, $B$, $C$ three real angles such that $A + B + C$ is a multiple of $180^{\circ}$. Then,
$ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$.
We will only show a proof of Theorem 2 using Theorem 1: First, we can WLOG assume that $A + B + C = 180^{\circ}$. This is because the inequality
$ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$
will not change if we add a multiple of 180° to one of the angles A, B and C (because $ \sin^{2}\left(180^{\circ} + u\right) = \sin^{2}u$ for every u), and consequently, since A + B + C is a multiple of 180°, we can add a multiple of 180° to the angle A such that, after this, we will have A + B + C = 180°.
Now, for A + B + C = 180°, we have
$ \left(180^{\circ} - 2A\right) + \left(180^{\circ} - 2B\right) + \left(180^{\circ} - 2C\right) = 540^{\circ} - 2\cdot\left(A + B + C\right)$ $ = 540^{\circ} - 2\cdot 180^{\circ} = 180^{\circ}$.
Hence, Theorem 1 (applied to $ 180^{\circ}-2A$, $ 180^{\circ}-2B$, $ 180^{\circ}-2C$ instead of $ A$, $ B$, $ C$) yields
$ x^{2} + y^{2} + z^{2}\geq 2yz\cos\left(180^{\circ} - 2A\right) + 2zx\cos\left(180^{\circ} - 2B\right) + 2xy\cos\left(180^{\circ} - 2C\right)$.
Since $ \cos\left(180^{\circ} - 2A\right) = - \cos\left(2A\right) = - \left(1 - 2\sin^{2}A\right) = 2\sin^{2}A - 1$ and similarly $ \cos\left(180^{\circ} - 2B\right) = 2\sin^{2}B - 1$ and $ \cos\left(180^{\circ} - 2C\right) = 2\sin^{2}C - 1$, this becomes
$ x^{2} + y^{2} + z^{2}\geq 2yz\left(2\sin^{2}A - 1\right) + 2zx\left(2\sin^{2}B - 1\right) + 2xy\left(2\sin^{2}C - 1\right)$ $ \Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right) - \left(2yz + 2zx + 2xy\right)$ $ \Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2} + \left(2yz + 2zx + 2xy\right)\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$ $ \Longleftrightarrow\ \ \ \ \ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$,
and Theorem 2 is proven.
Theorem 2 also trivially follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 and was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3849 ...