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.).
In http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=69200an ancient MathLinks topic (post #6) 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 measures is $9$ times the square of the distance between the point $P$ and the centroid of triangle $ABC$ ABC$. For better searchability, let me copy my MathLinks posts over here (up to finding some old post on MathLinks is almost impossible as for now). Note that I do not claim originality for the theorems. Theorem 1. Let$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 constant 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 factor)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)$$=360^{\circ}-\left(180^{\circ}-C\right)-\left(180^{\circ}-A\right)=C+A=180^{\circ}-B (since A + B + C = 180°). Now, all the vectors \overrightarrow{a}, \overrightarrow{b} and \overrightarrow{c} have unit length: \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1. Thus, \measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A yields \overrightarrow{b}\cdot\overrightarrow{c}=\left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|\cdot\cos\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=1\cdot 1\cdot\cos\left(180^{\circ}-A\right)=\cos\left(180^{\circ}-A\right)$$=-\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)$$= 540^{\circ} - 2\cdot 180^{\circ} = 180^{\circ}. Hence, Theorem 1 (applied to  180^{\circ}-A,  180^{\circ}-B,  180^{\circ}-C instead of  A,  B,  C) yields  x^{2} + y^{2} + z^{2}\geq 2yz\cos\left(180^{\circ} - 2A\right) + 2zx\cos\left(180^{\circ} - 2B\right) + 2xy\cos\left(180^{\circ} - 2C\right). Since  \cos\left(180^{\circ} - 2A\right) = - \cos\left(2A\right) = - \left(1 - 2\sin^{2}A\right) = 2\sin^{2}A - 1 and similarly  \cos\left(180^{\circ} - 2B\right) = 2\sin^{2}B - 1 and  \cos\left(180^{\circ} - 2C\right) = 2\sin^{2}C - 1, this becomes  x^{2} + y^{2} + z^{2}\geq 2yz\left(2\sin^{2}A - 1\right) + 2zx\left(2\sin^{2}B - 1\right) + 2xy\left(2\sin^{2}C - 1\right)$$ \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)$, and Theorem 2 is proven. Theorem 2 also trivially follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 and was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3849 ... 1 In http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=69200 (post #6) I have given a proof of the inequality by reducing it to$\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$. Then, the difference between the left hand side and the right hand side of your inequality measures the square of the distance between the point$P$and the centroid of triangle$ABC\$ (up to a constant scalar factor).