we know [Lagrange's identity](https://en.wikipedia.org/wiki/Lagrange%27s_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
$$(a^2\_{1}+a^2\_{2}+a^2\_{3})(b^2\_{1}+b^2\_{2}+b^2\_{3})\ge (a\_{1}b\_{1}+a\_{2}b\_{2}+a\_{3}b\_{3})^2$$


However, 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 $a\_{i},b\_{i}\in \mathbb R,i=1,2,3$?



An inequality concerning Lagrange's identity
Without loss of generality, all the $a\_i$'s and $b\_i$'s are nonzero. 
Let $\tilde d$ denote the difference between the left- and right-hand sides of the conjectured inequality $(\*)$, which then of course can be rewritten as $\tilde d\ge0$. 
In the previous version of my answer, I rewrote $\tilde d$ in new variables, $x\_i$ and $y\_i$, after which the inequality $\tilde d\ge0$ could be (rigorously) verified with Mathematica (in about 22 min). 


Here that expression for $\tilde d$ is further rewritten -- in new, "more-macro", variables -- so that the resulting expression can be rather easily analyzed, to prove the inequality $(\*)$. Indeed, let 
$p\_i:=(x\_i-y\_i)y\_i$, $x\_i:=a\_1 a\_2 a\_3/a\_i$, $y\_i:=b\_1 b\_2 b\_3/b\_i$, 
\begin{equation}
c\_1:=p\_2^2 + p\_2 p\_3 + p\_3^2\ge0,\quad c\_2:=p\_1^2 + p\_1 p\_3 + p\_3^2\ge0,\quad c\_3:=p\_2^2 + p\_2 p\_1 + p\_1^2\ge0, \tag{0} 
\end{equation}
and $z\_i:=y\_i^2\ge0$. 
Note that $x\_1 x\_2 x\_3=(a\_1 a\_2 a\_3)^2>0$ and 
$y\_1 y\_2 y\_3=(b\_1 b\_2 b\_3)^2>0$; moreover, 
\begin{equation}
(p\_1+z\_1)(p\_2+z\_2)(p\_3+z\_3)\ge0. \tag{1} 
\end{equation} 
The crucial identity is  

$$
 \tilde d\,y\_1 y\_2 y\_3=d:=
 p\_1 p\_2 p\_3+c\_1 z\_1+c\_2 z\_2+c\_3 z\_3. 
$$
Since $y\_1 y\_2 y\_3>0$, $\tilde d$ equals $d$ in sign. So, it suffices to show that $d\ge0$ -- for any real $p\_i$'s, the $c\_i$'s as in $(0)$, and any nonnegative $z\_i$'s satisfying $(1)$. 
Note here that without loss of generality $p\_1 p\_2 p\_3<0$ -- otherwise, $d\ge0$ immediately follows because the $c\_i$'s and $z\_i$'s are nonnegative. So, we may assume that the $p\_i$'s are are all nonzero and hence the $c\_i$'s are all strictly positive. 


Take any nonzero real $p\_i$'s and any nonnegative $z\_i$'s such that $(1)$ holds. Let us then fix those $z\_1$ and $z\_2$, and let $z\_3$ be decreasing as long as $z\_3$ remains nonnegative and $(1)$ holds; clearly, this process can stop only when the value of $z\_3$ becomes either $0$ or $-p\_3$, and in the latter case we must have $-p\_3>0$. 
Moreover, since $c\_i>0$ for all $i$, the value of $d$ will not increase after this process is complete. 


We can then proceed similarly by decreasing $z\_2$ (instead of $z\_3$), and then by decreasing $z\_1$. 


Let now $(z\_1,z\_2,z\_3)$ be any minimizer of $d$. Then it follows from the above reasoning that $z\_i\in\{0,-p\_i\}$ for each $i=1,2,3$; moreover, if at that $z\_i=-p\_i$ for some $i$, then we must have $-p\_i>0$. 
So, by the symmetry with respect to permutations of the indices, it is enough to consider the following four cases: 


(i) $z\_1=-p\_1>0$, $z\_2=-p\_2>0$, $z\_3=-p\_3>0$; 


(ii) $z\_1=-p\_1>0$, $z\_2=-p\_2>0$, $z\_3=0$; 


(iii) $z\_1=-p\_1>0$, $z\_2=0$, $z\_3=0$; 


(iv) $z\_1=0$, $z\_2=0$, $z\_3=0$, so that $(1)$ becomes $p\_1 p\_2 p\_3\ge0$. 


In case (i), $\min\_{z\_1,z\_2,z\_3}d=-(p\_1 + p\_2) (p\_1 + p\_3) (p\_2 + p\_3)>0$. 


In case (ii), $\min\_{z\_1,z\_2,z\_3}d=-p\_1 p\_2 (p\_1 + p\_2) - p\_1 p\_2 p\_3 + (-p\_1 - p\_2) p\_3^2$, which is a convex quadratic polynomial in $p\_3$, with discriminant $-p\_1 p\_2 (4 p\_1^2 + 7 p\_1 p\_2 + 4 p\_2^2)<0$, whence again $\min\_{z\_1,z\_2,z\_3}d>0$. 


In case (iii), $\min\_{z\_1,z\_2,z\_3}d=-p\_1 (p\_2^2 + p\_3^2)>0$. 


In case (iv), $\min\_{z\_1,z\_2,z\_3}d=p\_1 p\_2 p\_3\ge0$. 


Thus, $\min\_{z\_1,z\_2,z\_3}d\ge0$ in all cases, and the inequality in question is proved. 


