Inequality within a triangle

Let $ABC$ be a triangle. Prove that:

$$\prod \frac{\sin^2 A + \sin B \sin C}{\sin B + \sin C} \geq \frac{E}{2R^2}$$

where $E$ is the area of the triangle.

Solution

Let $x= \sin A, \; y = \sin B, \; z= \sin C$. Then we have to prove that:

$$\left ( x^2+yz \right )\left ( y^2+zx \right )\left ( z^2+xy \right )\geq xyz \left ( x+y \right )\left ( y+z \right )\left ( z+x \right )$$

Yeah, but this immediate from Cauchy - Schwartz inequality since:

$$\left ( x^2+yz \right )\left ( 1 + \frac{y}{z} \right )\geq \left ( x+y \right )^2 \Leftrightarrow \left ( x^2+yz \right )\left ( y+z \right )\geq z\left ( x+y \right )^2$$

Mutliplying cyclic the three inequalities that come up, yields the result.