Macedonian Mathematical Olympiad

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 $$a_i^5{a}_j^3+a_i^3{a}_j^5=a_i^3{a}_j^3(a_i^2+a_j^2)\ge a_i^3{a}_j^3(2a_i{a}_j)=2a_i^4{a}_j^4,\text{ for all }i<j.$$ 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 $$a_i^5{a}_j+a_i{a}_j^5=a_i{a}_j(a_i^4+a_j^4)\ge 2a_i{a}_j(a_i^2{a}_j^2)=2a_i^3{a}_j^3,\text{ for all }i<j.$$ It holds $t_5{t}_1\ge t_4{t}_2$, since $$(a_i^5{a}_j+a_i{a}_j^5)-(a_i^4{a}_j^2+a_i^2{a}_j^4)=a_i{a}_j(a_i-a_j{)}^2(a_i^2+a_i{a}_j+a_j^2)\ge 0.$$ $$t_5^4{t}_3^2{t}_1^2\ge t_5^2{t}_4^4{t}_1^2\ge t_5{t}_4^4{t}_3^2{t}_1\ge t_4^5{t}_3^2{t}_2,$$ so, we obtain $$t_5^4{t}_1^2\ge t_4^5{t}_2.$$