# If a,b,c,d,e are positive reals then prove the following inequality

$$\sum_{cyclic}\frac{a}{b+c}\geq\frac{5}{2}$$ 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!

• That is Shapiro's inequality for $n=5$. Commented Jun 28, 2021 at 9:27
• @Martin R Woah I am new to inequalities... Thanks for letting me know! Commented Jun 28, 2021 at 11:38

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 $$\sum_{sym}(a-b)^2\geq0$$