India — Team Selection Test, 2007

We use the standard
$$w_a = \frac{2bc \cos(A/2)}{(b+c)}, \text{ etc.}$$

The inequality takes the form
$$\sum_{\text{cyclic}} \frac{(b^2 + c^2)(b+c)}{4Rbc \cos(A/2)} > 2.$$
This may be put in the form
$$\sum_{\text{cyclic}} \frac{(b^2 + c^2)(b+c) \sin(A/2)}{2abc} > 1.$$
But note that $b^2+c^2 \ge 2bc$ and $b+c > a$. Hence it is sufficient to prove that $\sum_{\text{cyclic}} \sin(A/2) > 1$. We start with the identity
$$\sum_{\text{cyclic}} \cos A = 1 + 4 \prod_{\text{cyclic}} \sin(A/2),$$
which shows that $\sum_{\text{cyclic}} \cos A > 1$, in any triangle $ABC$. But, whenever $A, B, C$ are the angles of a triangle, we know that $(\pi - A)/2, (\pi - B)/2, (\pi - C)/2$ are also the angles of some other triangle. For this triangle, we get
$$\sum_{\text{cyclic}} \cos \left( \frac{\pi - A}{2} \right) > 1.$$
It follows that
$$\sum_{\text{cyclic}} \sin \frac{A}{2} > 1,$$
which is to be proved.