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)  ?  

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