Áp dụng bđt (x+y+z).(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9 , ta có (3 - ab + 3 - bc + 3 - ca ).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} \geq 9 \Rightarrow (9 - ab - bc - ca ).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} \geq 9 (1)Ta có a2+b2+c2 \geq ab + bc + ca \Rightarrow ab+bc+ca \leq 3 Từ (1) \Rightarrow (\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca} \geq \frac{9}{9 - ab - bc - ca } \geq \frac{3}{2}Dấu = xảy ra \Leftrightarrow a=b=c=1
Áp dụng bđt (x+y+z).(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq9 , ta có (3-ab+3-bc+3-ca).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq9 \Rightarrow(9-ab-bc-ca).(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq9 (1)Ta có a2+b2+c2\geqab+bc+ca\Rightarrowab+bc+ca\leq3 Từ (1)\Rightarrow(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}\geq\frac{9}{9-ab-bc-ca}\geq\frac{3}{2}Dấu = xảy ra \Leftrightarrow a=b=c=1