An inequality concerning Lagrange's identity 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
$$(a^2_{1}+a^2_{2}+a^2_{3})(b^2_{1}+b^2_{2}+b^2_{3})\ge (a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3})^2$$
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 $a_{i},b_{i}\in \mathbb R,i=1,2,3$?
 A: I tried hard to find counterexample but failed after 2 days' running of SageMath's program as below.
However two rules could be oberserved:


*

*for any given k > 0, there are many solutions satisfy (left - right) / right < k

*when left = right,
either   ai = bi for i = 1,2,3
or       ai = bj = bk = 0 and i+j+k = 6
It seems the inequality is right. But I don't have ability to prove it.
####  a  conjecture on MO ####
import time,random
for j in xrange(10000000001):
if j % 1000000 == 0:
    print 'j reach ',j, time.ctime()
v =  200 # max       
a1 = random.randint(0, v)   #  can be modified to Real 
a2 = random.randint(0, v)
a3 = random.randint(0, v)
b1 = random.randint(0, v)
b2 = random.randint(0, v)
b3 = random.randint(0, v)
a21  = a1^2
b22 =  b2^2
b23 =  b3^2
a22  = a2^2     
b21 =  b1^2
a23  = a3^2
C = (a21+b22+b23)*(a22+b23+b21)*(a23+b21+b22)
D = (b21+b22+b23)*(a1*b1+a2*b2+a3*b3)^2+(b1*a2*b3-b1*b2*a3)^2/2+(b1*b2*a3-a1*b2*b3)^2/2+(a1*b2*b3-b1*a2*b3)^2/2  
if C < D:
    print a1,a2,a3,b1,b2,b3,'counterexample!'
    break
if C > D and (C-D)/C < 0.000001: print C,D,1.0*(C-D)/C,'for', a1,a2,a3,b1,b2,b3
if C == D: print 'C=D for ', a1,a2,a3,b1,b2,b3,'should 3 pairs equal or three 0 in a special order'

print 'done'
A: This is a comment, not the answer. However it is too long as a comment.
Lagrange's identity follows from the properties of quaternions, namely that they constitute a composition algebra. Let $q_1=a_1i+a_2j+a_3k$ and $q_2=b_1i+b_2j+b_3k$ be two imaginary quaternions. Then $q_1q_2=-\vec{A}\cdot\vec{B}+(\vec{A}\times\vec{B})_1i+(\vec{A}\times\vec{B})_2j+(\vec{A}\times\vec{B})_4k$, where
$\vec{A}=(a_1,a_2,a_3)$ and $\vec{B}=(b_1,b_2,b_3)$. Lagrange's identity follows from the composition property of the quaternionic norm $|q_1q_2|=|q_1||q_2|$: $$\vec{A}^2\vec{B}^2=(\vec{A}\cdot\vec{B})^2+(\vec{A}\times\vec{B})^2.$$ Let us try to generalize Lagrange's identity by introducing third imaginary quaternion $q_3=c_1i+c_2j+c_3k$. Then 
$$q_1q_2q_3=-(\vec{A}\times\vec{B})\cdot\vec{C}+W_1i+W_2j+W_3k,$$ where $\vec{C}=(c_1,c_2,c_3)$ and $$\vec{W}=-\vec{A}\cdot\vec{B}\,\vec{C}+(\vec{A}\times\vec{B})\times\vec{C}=-\vec{A}\cdot\vec{B}\,\vec{C}-\vec{B}\cdot\vec{C}\,\vec{A}+\vec{C}\cdot\vec{A}\,\vec{B}.$$ Therefore it follows from $|q_1q_2q_3|=|q_1||q_2||q_3|$ that $$\vec{A}^2\vec{B}^2\vec{C}^2=[(\vec{A}\times\vec{B})\cdot\vec{C}]^2+$$ $$(\vec{A}\cdot\vec{B})^2\,\vec{C}^2+(\vec{B}\cdot\vec{C})^2\,\vec{A}^2+(\vec{C}\cdot\vec{A})^2\,\vec{B}^2-2(\vec{A}\cdot\vec{B})\,(\vec{B}\cdot\vec{C})\,(\vec{C}\cdot\vec{A}),$$ which can be represented also in the form $$\vec{A}^2\vec{B}^2\vec{C}^2=[(\vec{A}\times\vec{B})\times\vec{C}]^2+[(\vec{A}\times\vec{B})\cdot\vec{C}]^2+(\vec{A}\cdot\vec{B})^2\vec{C}^2. \tag{1}$$
Some non-trivial inequalities follow from this generalization of the Lagrange's identity, like $$\vec{A}^2\vec{B}^2\vec{C}^2\ge [(\vec{A}\times\vec{B})\cdot\vec{C}]^2-2(\vec{A}\cdot\vec{B})\,(\vec{B}\cdot\vec{C})\,(\vec{C}\cdot\vec{A}),$$ and $$\vec{A}^2\vec{B}^2\vec{C}^2\ge [(\vec{A}\times\vec{B})\times\vec{C}]^2+[(\vec{A}\times\vec{B})\cdot\vec{C}]^2,$$ however it is not clear how the inequality (*) is related to it (if related at all). Note that (1) simply follows from the Lagrange's identity itself. Therefore the considered generalization is not very profound. 
