$$\sum_{cyclic}\frac{a}{b+c}\geq\frac{5}{2}$$ My approach: I learned this technique in the book itself from which the question was taken, but doesn't quite seem to work. So,
Let, S=$\sum_{cyclic}\frac{a}{b+c}$ And we can write $$\sum_ca=\sum_c \frac {\sqrt a(\sqrt a (\sqrt {b+c}))}{\sqrt {b+c}}$$
On Applying the Cauchy Schwarz Inequality, $$\left(\sum_ca\right)^2\leq S\left(\sum_c a(b+c)\right)$$ All that remains for me to prove is that, $$\left(\sum_ca\right)^2\div\left(\sum_c a(b+c)\right) \geq \frac{5}{2}$$ I tried to prove this using the A.M.-G.M Inequality but that only helped me complicate matters. Also, I would I like to know if the cyclic sum is distributive over its elements. Thanks!