$$\sum_{cyclic}\frac{a}{b+c}\geq\frac{5}{2}$$
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!
Best Answer
By C-S $$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d+e)^2}{\sum\limits_{cyc}(ab+ac)}\geq\frac{5}{2},$$ where the last inequality it's just $$\sum_{sym}(a-b)^2\geq0$$