An inequality concerning Lagrange's identity 
Does the following inequality still hold
  $$(a^2_{1}+b^2_{2}+b^2_{3})(a^2_{2}+b^2_{3}+b^2_{1})(a^2_{3}+b^2_{1}+b^2_{2})\ge (b^2_{1}+b^2_{2}+b^2_{3})(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2 $$
  $$+\dfrac{1}{2}(b_{1}a_{2}b_{3}-b_{1}b_{2}a_{3})^2+\dfrac{1}{2}(b_{1}b_{2}a_{3}-a_{1}b_{2}b_{3})^2+\dfrac{1}{2}(a_{1}b_{2}b_{3}-b_{1}a_{2}b_{3})^2\tag{*}$$
  for $a_{i},b_{i}\in \mathbb R,i=1,2,3$?

we  know Lagrange's identity 
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})=(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2+\sum_{i=1}^{2}\sum_{j=i+1}^{3}(a_{i}b_{j}-a_{j}b_{i})^2$$
then we have Cauchy-Schwarz inequality
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$

We have the following identity.
$$(a^2+y^2+z^2)(b^2+x^2+z^2)(c^2+x^2+y^2)-(x^2+y^2+z^2)(ax+by+cz)^2-$$
$$-\frac{1}{2}(ayz-bxz)^2-\frac{1}{2}(ayz-cxy)^2-\frac{1}{2}(bxz-cxy)^2=$$
$$=(x^2+y^2)(x^2+z^2)(y^2+z^2)+a^2b^2c^2+a^2b^2(x^2+y^2)+a^2c^2(x^2+z^2)+b^2c^2(y^2+z^2)-$$
$$-2abxy(x^2+y^2)-2acxz(x^2+z^2)-2bcyz(y^2+z^2)-xyz(abz+acy+bcx).$$
I hope it can help.
