Write $\sigma_1 = a + b + c$, $\sigma_2 = ab + bc + ca$, $\sigma_3 = abc$. Then the inequality we have to prove becomes
$$\sigma_1^5 \ge 81(\sigma_1^2 - 2\sigma_2)\sigma_3 \iff \sigma_1^5 + 162\sigma_2\sigma_3 \ge 81\sigma_1^2\sigma_3. \quad (1)$$

Now using the AM-GM inequality for three summands $\sigma_1^5$, $81\sigma_2\sigma_3$, $81\sigma_2\sigma_3$ we get
$$\sigma_1^5 + 81\sigma_2\sigma_3 + 81\sigma_2\sigma_3 \ge 3\sqrt[3]{81^2\sigma_1^5\sigma_2^2\sigma_3^2}.$$
So for (1) to be proved it suffices to verify that
$$3\sqrt[3]{81^2\sigma_1^5\sigma_2^2\sigma_3^2} \ge 81\sigma_1^2\sigma_3 \iff 81^2\sigma_1^5\sigma_2^2\sigma_3^2 \ge 27^3\sigma_1^6\sigma_3^3 \iff \sigma_2^2 \ge 3\sigma_1\sigma_3.$$
The last inequality is well known.