Ireland_2017

$$4s_5{s}_1^3+s_2^4\ge 5s_4{s}_2{s}_1^2.$$ Using the AM-GM inequality, $4s_5{s}_1^3+s_2^4\ge 5[s_5{s}_1^{12}{s}_2^4{]}^{1/5}$, so it suffices to prove that $$s_5^4{s}_2^4{s}_1^{12}\ge s_4^5{s}_2^5{s}_1^{10},$$ and thus, using the positivity of $s_k$, that $s_5{s}_1^2\ge s_4{s}_2$.

Observe that $s_5{s}_3\ge s_4^2$, since, on expansion ${\sum }_{i=1}^n{a}_i^8$ occurs on both sides, and each term $(a_i^5{a}_j^3+a_i^3{a}_j^5)=(a_i^3{a}_j^3)(a_i^2+a_j^2)\ge a_i^3{a}_j^3(2a_i{a}_j)=2a_i^4{a}_j^4$ for all $i<j$.

Observe also that $s_5{s}_1\ge s_3^2$, since, on expansion, ${\sum }_{i=1}^n{a}_i^8$ occurs on both sides, and each term $(a_i^5{a}_j+a_i{a}_j^5)=a_i{a}_j(a_i^4+a_j^4)\ge 2a_i{a}_j(a_i^2{a}_j^2)=2a_i^3{a}_j^3$, $(i<j)$.

Observe further that $s_5{s}_1\ge s_4{s}_2$, since $$(a_i^5{a}_j+a_i{a}_j^5)-(a_i^4{a}_j^2+a_i^2{a}_j^4)=a_i{a}_j(a_i-a_j{)}^2(a_i^2+a_i{a}_j+a_j^2)\ge 0.$$ Now $s_5{s}_3^2{s}_1^2\ge s_2^2{s}_4{s}_1^2\ge s_5{s}_4{s}_3^2{s}_1\ge s_4^5{s}_2^2{s}_2$, and the result follows.