notice for (a) it is equal to $12-2(ab+bc+ca)$ $(a^2+b^2+c^2) = 6 $ by cauchy $-3\sqrt{2} \leq(a+b+c) \leq 3\sqrt{2}$ $ab+bc+ca = \frac{(a+b+c)^2-a^2-b^2-c^2}{2}=\frac{(a+b+c)^2-6}{2}$ $2ab+2bc+2ca=(a+b+c)^2-6$ $12-2ab-2bc-2ca = 18-(a+b+c)^2 \leq 18$ Equality cases $a+b+c =0$ $a^2+b^2+c^2=6$ $ab+bc+ca=-3$ $bc = a^2+3$ $b+c=-a$ $f(b)=b^2+ab+(a^2-3)=0$ (b,c) roots of eqn

Let $a, b, c$ be reals satisfying $a^2+b^2+c^2=6$. Prove that $$(a-b)^2+(b-c)^2+(c-a)^2\leq 18$$$$(a-b)^2 (b-c)^2 (c-a)^2\leq 108$$

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Let's say the A to this polynom. $ab+ac+bc=((a+b+c)^2-a^2-b^2-c^2)/2$ For $(a+b+c)^2$, we can get this inequality; $3\sqrt{2}>=(a+b+c)^2>=0$ If we make it bigger ( A polynom), so we must decrease $(a+b+c)^2$. Because, $a^2+b^2+c^2$ is constant and it's given us. Then $ab+ac+bc=-3$ $A<=2*(a^2+b^2+c^2)-2*(ab+ac+bc)=12+6=18$