The inequality that we need to prove is equivalent with the inequality
$$2t_5^2 t_1^6 - 5t_4^4 t_2^2 t_1^2 + 3t_2^3 t_4^5 \ge 0,$$
or with
$$2t_5^2 t_1^6 + 3t_2^3 t_4^5 \ge 5t_4^4 t_2^2 t_1^2.$$
Using the inequality between arithmetic and geometric mean we have
$$2t_5^2 t_1^6 + 3t_2^3 t_4^5 \ge 5(t_5^4 t_1^{12} t_2^9 t_4^{15})^{\frac{1}{5}}.$$
By the last inequality it is enough to prove that
$$t_5^4 t_1^{12} t_2^9 t_4^{15} \ge t_4^{20} t_2^{10} t_1^{10},$$
i.e.
$$t_5^4 t_1^2 \ge t_4^5 t_2.$$
Let us note that $t_5 t_3 \ge t_4^2$, since $\sum_{i=1}^n a_i^8$ is appearing on the both sides of the inequality and moreover holds
$$a_i^5 a_j^3 + a_i^3 a_j^5 = a_i^3 a_j^3 (a_i^2 + a_j^2) \ge a_i^3 a_j^3 (2a_i a_j) = 2 a_i^4 a_j^4, \text{ for all } i < j.$$
Furthermore, it holds $t_5 t_1 \ge t_3^2$. Indeed, $\sum_{i=1}^n a_i^6$ is appearing on the both sides of the inequality and moreover
$$a_i^5 a_j + a_i a_j^5 = a_i a_j (a_i^4 + a_j^4) \ge 2 a_i a_j (a_i^2 a_j^2) = 2 a_i^3 a_j^3, \text{ for all } i < j.$$
It holds $t_5 t_1 \ge t_4 t_2$, since
$$(a_i^5 a_j + a_i a_j^5) - (a_i^4 a_j^2 + a_i^2 a_j^4) = a_i a_j (a_i - a_j)^2 (a_i^2 + a_i a_j + a_j^2) \ge 0.$$
$$t_5^4 t_3^2 t_1^2 \geq t_5^2 t_4^4 t_1^2 \geq t_5 t_4^4 t_3^2 t_1 \geq t_4^5 t_3^2 t_2,$$
so, we obtain
$$t_5^4 t_1^2 \geq t_4^5 t_2.$$