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}________________________________$=\frac{
8}{9}$______________________