Hello IMOment 3

The equation a²+b²+c²=1 is not a homogeneous expression in a,b,c, thus we can put (a,b,c)=( $\frac {x}{\sqrt{x²+y²+z²}}$ , $\frac {y}{\sqrt{x²+y²+z²}}$ , $\frac {z}{\sqrt{x²+y²+z²}}$ ) to satisfy it. Thus our expression (a²-bc)(b²-ca)(c²-ab) becomes $\frac {(x²-yz)(y²-zx)(z²-xy)}{(x²+y²+z²)³}$ . Now we notice that, (x+y+z)²>=0 =>3(x²+y²+z²)>=2{(x²-yz)+(y²-zx)+(z²-xy)}. Equality holds when x+y+z=0, but then
(x-y)(x+y+z)=0 => x²-yz=y²-zx...and so on.
But if a=b=c then a+b+c=3 $\sqrt[3]{abc}$ . Thus we have,
$\frac{3}{2}$ (x²+y²+z²)>=3 $\sqrt[3]{(x²-yz)(y²-zx)(z²-xy)}$ by combining our previous inequalities. Thus we can cube the inequality and divide it by (x²+y²+z²)³ and see that, (a²-bc)(b²-ca)(c²-ab)<=⅛. Equality holds when a+b+c=0.

Let $k=a+b+c, |k|\leq \sqrt{3}$ . Then,

$\left(a^2-b c\right) \left(b^2-a c\right) \left(c^2-a b\right)==\left(1-\left(a^2+a b+b^2\right)\right) \left(1-\left(a^2+a c+c^2\right)\right) \left(1-\left(b^2+b c+c^2\right)\right)==\left(1-\frac{1}{2} \left(-a^2+(b+c)^2+1\right)\right) \left(1-\frac{1}{2} \left((a+c)^2-b^2+1\right)\right) \left(1-\frac{1}{2} \left((a+b)^2-c^2+1\right)\right)==\frac{1}{8} ((a-b-c) (a+b+c)+1) ((-a+b-c) (a+b+c)+1) (1-(a+b-c) (a+b+c))==\frac{1}{8} (k (2 a-k)+1) (k (2 b-k)+1) (k (2 c-k)+1)\leq \frac{1}{8} \left(\frac{1}{3} \left(3-k^2\right)\right)^3\leq \frac{1}{8}$

where we used AM-GM. It is easy to find an example where the maximum is achieved.