Note that if the line $ L$ lies entirely outside the triangle, we can decrease each of $ u,v,w$ by sliding $ L$, keeping it parallel to its initial position, until it passes through one of the vertices. Therefore we need only consider lines that pass through the interior (including border) of the triangle. Of those lines, we can further slide in a parallel fashion and decrease the sum of $ u,v,w$ until the setup is equivalent to: $ A$ and $ B$ lie on opposite sides of $ L$, $ L$ passes through $ C$. So relabel diagram if necessary to this. This is true because $ u+v$ remains constant as we slide around iff $ B$ and $ C$ are on opposite sides of $ L$. Meanwhile $ w$ decreases until it is zero, the minimum, which is just when $ L$ is through $ A$. Back to the inequality. We can show it is equivalent to \[ \frac{u^2}{b\cos A}+\frac{v^2}{b\cos A}\geq c\] If $ u=b\sin( A+X)$ then $ v=a\sin (B-X)$ So we need show \[ b\frac{\sin^2(A+X)}{\cos A}+a\frac{\sin^2(B-X)}{\cos B}\geq c\] We will verify by checking boundary cases as well as zeros of deriv. (as function of $ X$) The deriv. is zero when: \[ b\frac{\sin 2(A+X)}{\cos A} = a\frac{\sin 2(B-X)}{\cos B}\] \[ \iff\frac{\sin 2(A+X)}{\sin 2A} = \frac{\sin 2(B-X)}{\sin 2B}\] \[ \iff \cos 2(A + X - B) - \cos 2(A+X+B) = \cos 2(B-X-A) - \cos 2(A+B-X) \] \[ \iff \cos 2(A+B+X) = \cos 2(A+B-X) \iff \sin 2(A+B) \sin X = 0\] Thus we need only check boundaries; i.e., $ X=0,\frac{\pi}{2}$; but at both we clearly see equality attained ($ a\cos B + b\cos A = c$) So we have proven inequality with equality when $ L$ is one of the sides of the triangle.

me@home wrote: This is true because $ u + v$ remains constant as we slide around iff $ B$ and $ C$ are on opposite sides of $ L$. Meanwhile $ w$ decreases until it is zero, the minimum, which is just when $ L$ is through $ A$. Sorry to revive such an old topic, but I don't quite see how $ u+v$ is constant implies the minimum of the expression occurs when $ w$ is minimum. Can someone elaborate on me@home's solution?