Homogeneous fourth degree inequality : $\sum x_i^2x_j^2 +6x_1x_2x_3x_4 \geq\sum x_ix_jx_k^2$

Question

Let $x_1,x_2,x_3,x_4$ be four real numbers. The inequality

$$ \sum_{i,j} x_i^2x_j^2 +6x_1x_2x_3x_4 \geq \sum_{i,j,k} x_ix_jx_k^2 $$ (where the sums are over unordered uples of distinct indices) is true because if one views the difference as a trinomial in $x_4$, this trinomial has negative discriminant $\frac{-3}{16}((x_2-x_1)(x_3-x_1)(x_3-x_2))^2$. The drawback of this method are (1) it is not very symmetrical, (2) it involves the somewhat painful computation of the discriminant, and (3) the equality case is not very easy to deduce from it.

Are there other methods, improving on (1) and/or (2) and/or (3) ?

Answer 1

It's $$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2\geq0.$$ Because if $x_1=a$, $x_2=b$, $x_3=c$ and $x_4=d$ then:

$$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2=$$ $$=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2=$$ $$=(a^2-2ab+b^2)(c^2-2cd+d^2)+$$ $$+(a^2-2ac+c^2)(b^2-2bd+d^2)+$$ $$+(a^2-2ad+d^2)(b^2-2bc+c^2)=$$ $$=a^2c^2+a^2d^2+b^2c^2+b^2d^2-2a^2cd-2b^2cd-2c^2ab-2d^2ab+4abcd+$$ $$+a^2b^2+a^2d^2+b^2c^2+c^2d^2-2a^2bd-2c^2bd-2b^2ac-2d^2ac+4abcd+$$ $$+a^2b^2+a^2c^2+b^2d^2+c^2d^2-2a^2bc-2d^2bc-2b^2ad-2c^2ad+4abcd=$$ $$=2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)+12abcd-2\sum_{cyc}a^2(bc+bd+cd).$$

Answer 2

Perhaps another approach is more understandable. Let $$q_1 :=(x_1x_2)^2+(x_1x_3)^2+(x_1x_4)^2+(x_2x_3)^2+(x_2x_4)^2+(x_3x_4)^2,\quad q_2 := 6x_1x_2x_3x_4$$ and $$q_3 :=(x_1x_2+x_1x_3+x_2x_3)x_4^2+(x_1x_2+x_1x_4+x_2x_4)x_3^2 +(x_1x_3+x_1x_4+x_3x_4)x_2^2+(x_2x_3+x_2x_4+x_3x_4)x_1^2.$$ We could like to express $\;q_4 := q_1+q_2-q_3\;$ as a sum of squares. As a first step we notice that $q_5 := (x_1x_2+x_3x_4)^2+(x_1x_3+x_2x_4)^2+(x_1x_4+x_2x_3)^2 = q_1+q_2.\;$ So now we want to express $\;q_4 := q_5-q_3\;$ as a sum of squares. We notice that $$q_4 = x_1^2x_2^2 + x_1^2x_3^2 + x_2^2x_3^2+x_1^2x_4^2 + x_2^2x_4^2 +x_3^2x_4^2 + 6x_1x_2x_3x_4 + \cdots.$$ Noticing that $q_4$ is invariant if the $x_1,x_2,x_3,x_4$ are incremented by the same quantity, we find $$q_4 - (x_1-x_2)^2(x_3-x_4)^2 = (x_1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4),$$ and incrementing indices cyclically, $$q_4 - (x_2-x_3)^2(x_4-x_1)^2 = (x_2-x_4)(x_2-x_1)(x_3-x_4)(x_3-x_1).$$ Adding both these equations together gives $$ 2q_4 - (x_1-x_2)^2(x_3-x_4)^2 -(x_2-x_3)^2(x_4-x_1)^2 =(x_1-x_3)^2(x_2-x_4)^2.$$ Rearranging the terms gives us our wanted sum of squares.