$\left(a^3+b^3+c^3-3abc \right)^2=\left(\sum a(a^2-bc) \right)^2$ $\leq (a^2+b^2+c^2)\sum (a^2-bc)^2=(\sum a^2)\left((\sum a^2)^2-(\sum ab)^2 \right)^2 \leq 8$

trungdung19122002 wrote: Given $a;b;c$ satisfying $a^{2}+b^{2}+c^{2}=2$ . Prove that: a) $\left | a+b+c-abc \right |\leqslant 2$ . b) $\left | a^{3}+b^{3}+c^{3}-3abc \right |\leqslant 2\sqrt{2}$

luofangxiang wrote: trungdung19122002 wrote: Given $a;b;c$ satisfying $a^{2}+b^{2}+c^{2}=2$ . Prove that: a) $\left | a+b+c-abc \right |\leqslant 2$ . b) $\left | a^{3}+b^{3}+c^{3}-3abc \right |\leqslant 2\sqrt{2}$ Old.

Given $a;b;c$ satisfying $a^{2}+b^{2}+c^{2}=2$ . Prove that$$\left | a^{3}+b^{3}+c^{3}-abc \right |\leqslant 2\sqrt{2}$$

$a,b,c : a^2+b^2+c^2=t^2$ $0 \leq n \leq 3$ $\Rightarrow \left |a^3+b^3+c^3-nabc \right | \leq t^3$

dalarin01 wrote: $a,b,c : a^2+b^2+c^2=t^2$ $0 \leq n \leq 3$ $\Rightarrow \left |a^3+b^3+c^3-nabc \right | \leq t^3$ $(a^3+b^3+c^3-nabc)^2=(\sum a(a^2-\frac{n}{3}bc))^2\leq(\sum a^2)(\sum(a^2-\frac{n}{3}bc)^2)$ $= t^2((\sum a^2)^2-\frac{n}{3}(\sum ab)^2+\frac{(n-3)(n+6)}{9}(\sum a^2b^2))$ $\leq t^6$