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

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) ?

• If $x_1=x_2=x_3=1,x_4=-1$ then $6+(-6)$ is not $\ge 4$. – Somos Feb 19 '18 at 16:20
• @Somos You are mistaken. With your choice of values, the RHS is $0$ not $4$ (in fact, we have equality when three of the variables are equal). You can check it : there are twelve terms of the forms $x_ix_jx_k^2$. Three of them (when $k=4$) are equal to $+1$, three others (when $k\neq 4$ and $i\neq 4,j\neq 4$) are equal to $+1$ and the rest (when $i=4$ or $j=4$) are equal to $-1$. – Ewan Delanoy Feb 19 '18 at 16:31
• Okay, let $x_1=x_2=x_3=x_4=1$. Then the first sum has $4*3=12$ addends each of $1$. The second term is $6$, and so the left side is $12+6=18$. The right side sum has $4*3*2=24$ addends each of $1$ so right side is $24$. – Somos Feb 19 '18 at 16:38
• @Somos You're wrong again on both sums. The first sum has six addends (six unordered pairs), while the RHS has 12 addends. I think you are confused because I forget to explicitly mention "unordered" indices in my OP. I added it now, thanks for your contribution. – Ewan Delanoy Feb 19 '18 at 16:51
• Okay. We agree on the 1st sum having 4 choose 2 = 6 addends. The right side sum has 4 choose 3 = 4 addends according to "unordered tuples". You need to be more explicit here. – Somos Feb 19 '18 at 16:59

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:
$$(x_1-x_2)^2(x_3-x_4)^2+(x_1-x_3)^2(x_2-x_4)^2+(x_1-x_4)^2(x_2-x_3)^2=$$ $$=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2=$$ $$=(a^2-2ab+b^2)(c^2-2cd+d^2)+$$ $$+(a^2-2ac+c^2)(b^2-2bd+d^2)+$$ $$+(a^2-2ad+d^2)(b^2-2bc+c^2)=$$ $$=a^2c^2+a^2d^2+b^2c^2+b^2d^2-2a^2cd-2b^2cd-2c^2ab-2d^2ab+4abcd+$$ $$+a^2b^2+a^2d^2+b^2c^2+c^2d^2-2a^2bd-2c^2bd-2b^2ac-2d^2ac+4abcd+$$ $$+a^2b^2+a^2c^2+b^2d^2+c^2d^2-2a^2bc-2d^2bc-2b^2ad-2c^2ad+4abcd=$$ $$=2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)+12abcd-2\sum_{cyc}a^2(bc+bd+cd).$$
• I used the following. Let $\sum\limits_{i=1}^4x_i=4u$,$\sum\limits_{1\leq i<j\leq4}x_ix_j=6v^2$, $\sum\limits_{1\leq i<j<k\leq4}x_ix_jk_k=4w^3$ and $\prod\limits_{i=1}^4x_i=t^4.$ Thus, we need to prove that $3u^4-4uw^3+t^4\geq0,$ which is my old inequality and I know that it's exactly, which I wrote above. – Michael Rozenberg Feb 19 '18 at 19:33
• Are you sure ? Unless I missed something, when you expand your LHS and divide out by 2 you get only $+1$ and $-1$ coeffs ,with no $6$ coeff as in my OP – Ewan Delanoy Feb 19 '18 at 19:33
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.