An inequality with four variables Prove that $$4[(4;0;0;0)]+12[(2;1;1;0)]\ge 12[(3;1;0;0)]+3[(1;1;1;1)]$$ with $$[(a;b;c;d)]=\frac{1}{4!}\sum_{\sigma\in Sym(4)}x_{\sigma(1)}^{a}\cdot x_{\sigma(2)}^b\cdot x_{\sigma(3)}^c\cdot x_{\sigma(4)}^d$$ Eh , the problem is that I need to show that $$(x_1−x_2)(x_1−x_3)(x_1−x_4)(x_1−x_5)+(x2−x1)(x2−x3)…$$ $$(x2−x−5)+…+(x5−x1)…(x5−x4)\geq0,$$ where $x_i$ are real numbers First of all , I see that ($x1;x2;x3;x4;x5)$ verify the inequlity then also $(x1+k;x2+k;x3+k;x4+k;x5+k)$ so we may assume that $x_5=\min(x_i)$ and let $k=−x_5$ so all the variables are gonna be postives and we will have only 4 variables , the I had two choices either developping (which is the problem I post) or assuming that $x_1=1$, but if I did so the inequality won't be homogenized

We need to prove that $$\sum_{cyc}(a-b)(a-c)(a-d)(a-e)\geq0.$$ Let $a\geq b\geq c\geq d\geq e$.

Thus, $$(c-a)(c-b)(c-d)(c-e)\geq0,$$ $$(a-b)(a-c)(a-d)(a-e)+(b-a)(b-c)(b-d)(b-e)=$$ $$=(a-b)((a-c)(a-d)(a-e)-(b-e)(b-c)(b-d))\geq0$$ and $$(d-a)(d-b)(d-c)(d-e)+(e-a)(e-b)(e-c)(e-d)=$$ $$=(d-e)((a-e)(b-e)(c-e)-(a-d)(b-d)(c-d))\geq0$$ and we are done!