bđt côsi ta có :$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \geqslant (a+b+c)(ab+bc+ca)-\frac{1}{9}$________________________________$=\frac{8}{9}$______________________
bđt côsi ta có :$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \geqslant (a+b+c)(ab+bc+ca)-\frac{1}{9}$________________________________$=\frac{1}{9}$______________________
bđt côsi ta có :$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \geqslant (a+b+c)(ab+bc+ca)-\frac{1}{9}$________________________________$=\frac{
8}{9}$______________________