Ta có: \((1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ca+abc\)Ta lại có: \(\left\{
\begin{array}{l}a+b+c\geq 3\sqrt[3]{abc}\qquad (1)\\ab+bc+ca\geq
3\sqrt[3]{(abc)^{2}}\qquad (2)\\abc=\sqrt[3]{(abc)^{3}} \end{array} \right. \);Do áp dụng BĐT Cauchy cho 3 số $a;b;c\ge 0\Rightarrow (1)$ $ab;bc;ca\ge 0\Rightarrow (2)$Vậy: \((1+a)(1+b)(1+c)\geq 1+3\sqrt[3]{(abc)}+3\sqrt[3]{(abc)^2}+\sqrt[3]{(abc)^3}\\\Leftrightarrow (1+a)(1+b)(1+c)\ge(1+\sqrt[3]{abc})^{3}\).Vậy ta có đpcm.Dấu bằng xảy ra khi$\left\{ \begin{array}{l} a=b=c\\ ab=bc=ca \end{array} \right.\Rightarrow a=b=c.$
Ta có: \((1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ca+abc\)Ta lại có: \(\left\{
\begin{array}{l}a+b+c\geq 3\sqrt[3]{abc}\qquad (1)\\ab+bc+ca\geq
3\sqrt[3]{(abc)^{2}}\qquad (2)\\abc=\sqrt[3]{(abc)^{3}} \end{array} \right. \);Do áp dụng BĐT Cauchy cho 3 số $a;b;c\ge 0\Rightarrow (1)$ $ab;bc;ca\ge 0\Rightarrow (2)$Vậy: \((1+a)(1+b)(1+c)\geq 1+\sqrt[3]{(abc)}+\sqrt[3]{(abc)^2}+\sqrt[3]{(abc)^3}\\\Leftrightarrow (1+a)(1+b)(1+c)\ge(1+\sqrt[3]{abc})^{3}\).Vậy ta có đpcm.Dấu bằng xảy ra khi$\left\{ \begin{array}{l} a=b=c\\ ab=bc=ca \end{array} \right.\Rightarrow a=b=c.$
Ta có: \((1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ca+abc\)Ta lại có: \(\left\{
\begin{array}{l}a+b+c\geq 3\sqrt[3]{abc}\qquad (1)\\ab+bc+ca\geq
3\sqrt[3]{(abc)^{2}}\qquad (2)\\abc=\sqrt[3]{(abc)^{3}} \end{array} \right. \);Do áp dụng BĐT Cauchy cho 3 số $a;b;c\ge 0\Rightarrow (1)$ $ab;bc;ca\ge 0\Rightarrow (2)$Vậy: \((1+a)(1+b)(1+c)\geq 1+
3\sqrt[3]{(abc)}+
3\sqrt[3]{(abc)^2}+\sqrt[3]{(abc)^3}\\\Leftrightarrow (1+a)(1+b)(1+c)\ge(1+\sqrt[3]{abc})^{3}\).Vậy ta có đpcm.Dấu bằng xảy ra khi$\left\{ \begin{array}{l} a=b=c\\ ab=bc=ca \end{array} \right.\Rightarrow a=b=c.$