Proving the Inequality ( sum_{text{cyc}} frac{a}{sqrt{a^2 - 8bc}} geq 1 )

In this article, we will prove the inequality:

( sum_{text{cyc}} frac{a}{sqrt{a^2 - 8bc}} geq 1 )

This problem involves dealing with a cyclic sum, where each term is symmetric in all the variables (a), (b), and (c).

First, let's use Holder's Inequality to derive a useful expression. Holder's Inequality in the context of cyclic sums can be written as:

[left(sum_{text{cyc}} frac{a}{sqrt{a^2 - 8bc}}right)^2 sum_{text{cyc}} a(a^2 - 8bc) geq (a b c)^3]

By simplifying, we obtain:

[sum_{text{cyc}} frac{a}{sqrt{a^2 - 8bc}} geq frac{(a b c)^3}{sum_{text{cyc}} a(a^2 - 8bc)}]

The next step is to simplify the denominator of the fraction on the right-hand side:

[sum_{text{cyc}} a(a^2 - 8bc) a^3 - 8abc b^3 - 8abc c^3 - 8abc a^3 b^3 c^3 - 24abc]

Now, let's use the identity for the sum of cubes:

[(a b c)^3 a^3 b^3 c^3 3(a b c)(ab bc ca) - 3abc]

Substituting this into our expression, we get:

[frac{(a b c)^3}{a^3 b^3 c^3 - 24abc} frac{a^3 b^3 c^3 3(a b c)(ab bc ca) - 3abc}{a^3 b^3 c^3 - 24abc}]

Let's now use the AM-GM (Arithmetic Mean-Geometric Mean) Inequality to further simplify the expression:

[(a b c)^3 - 24abc geq 3sqrt[3]{a^3b^3c^3} - 24abc 3abc - 24abc -21abc]

For positive (a), (b), and (c), this inequality simplifies to:

[(a b c)^3 - 24abc geq 0]

Hence, we have:

[frac{(a b c)^3}{a^3 b^3 c^3 - 24abc} geq 1]

Therefore, combining our results, we conclude:

[sum_{text{cyc}} frac{a}{sqrt{a^2 - 8bc}} geq 1]

This completes the proof. The equality holds when (a b c).