an M that drive you crazy

The left-hand side is equal to $|(a-b)(b-c)(c-a)(a+b+c)|$ . Since the inequality is homogeneous, $M$ the is maximum value of $|(a-b)(b-c)(c-a)(a+b+c)|$ subject to the constraint $a^2 + b^2 + c^2 = 1$ . Writing $X = \tfrac{1}{\sqrt{2}}(a-b) \hspace{2cm} Y \; = \; \tfrac{1}{\sqrt{6}}(a + b - 2c) \hspace{2cm} Z \; = \; \tfrac{1}{\sqrt{3}}(a + b + c)$ we see that $M$ is the maximum value of $\sqrt{\tfrac32}|X(3Y^2-X^2)Z|$ subject to $X^2 + Y^2 + Z^2 = 1$ . If we introduce polar coordinates $X = \sin\theta\cos\phi$ , $Y = \sin\theta\sin\phi$ , $Z = \cos\theta$ , then $M$ is the maximum of $\sqrt{\tfrac32} \big|\sin^3\theta \cos\theta \cos\phi(3 - 4\cos^2\phi)\big|$ Simple calculus tells us that the maximum of $|\sin^3\theta\cos\theta|$ for $0 \le \theta \le \pi$ is $\tfrac{3}{16}\sqrt{3}$ (when $\theta = \tfrac13\pi$ ), and the maximum value of $\cos\phi(3 - 4\cos^2\phi)$ for $0 \le \phi \le 2\pi$ is $1$ (when $\phi = \tfrac13\pi$ ). Thus $M = \boxed{\tfrac{9}{16\sqrt{2}}}$