giả sử $a\geq b\geq c$ khi đó$b^{2}-bc+c^{2}=b^{2} +c(c-b)\leq b^{2}$$c^{2}-ac+c^{2}=c(c-a)+a^{2} \leq a^{2}$$\Rightarrow P\leq a^{2}b^{2}(a^{2}-ab+b^{2})=\frac{4}{9}.\frac{3ab}{2}.\frac{3ab}{2}(a^{2}-ab+b^{2})$$\leq \frac{4}{9}( \frac{\frac{3ab}{2}+\frac{3ab}{2}+(a^{2}-ab+b^{2})}{3})^{3}=\frac{4}{9.27}(a+b)^{6}\leq \frac{4}{9.27}(a+b+c)^{6}=12$Dấu "=" $ a=2;b=1;c=0$
giả sử $a\geq b\geq c$ khi đó$b^{2}-bc+c^{2}=b^{2} +c(c-b)\leq b^{2}$$c^{2}-ac+c^{2}=c(c-a)+a^{2} \leq a^{2}$$\Rightarrow P\leq a^{2}b^{2}(a^{2}-ab+b^{2})=\frac{4}{9}.\frac{3ab}{2}.\frac{3ab}{2}(a^{2}-ab+b^{2})$$\leq \frac{4}{9}( \frac{\frac{3ab}{2}+\frac{3ab}{2}.(a^{2}-ab+b^{2})}{3})^{3}=\frac{4}{9.27}(a+b)^{6}\leq \frac{4}{9.27}(a+b+c)^{6}=12$Dấu "=" $ a=2;b=1;c=0$
giả sử $a\geq b\geq c$ khi đó$b^{2}-bc+c^{2}=b^{2} +c(c-b)\leq b^{2}$$c^{2}-ac+c^{2}=c(c-a)+a^{2} \leq a^{2}$$\Rightarrow P\leq a^{2}b^{2}(a^{2}-ab+b^{2})=\frac{4}{9}.\frac{3ab}{2}.\frac{3ab}{2}(a^{2}-ab+b^{2})$$\leq \frac{4}{9}( \frac{\frac{3ab}{2}+\frac{3ab}{2}
+(a^{2}-ab+b^{2})}{3})^{3}=\frac{4}{9.27}(a+b)^{6}\leq \frac{4}{9.27}(a+b+c)^{6}=12$Dấu "=" $ a=2;b=1;c=0$