Ta có: $3\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)={{\left( a+b+c \right)}^{2}}$
$\Leftrightarrow $$3{{a}^{2}}+3{{b}^{2}}+3{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$

$\Leftrightarrow $$({{a}^{2}}-2ab+{{b}^{2}})+({{b}^{2}}-2bc+{{c}^{2}})+({{c}^{2}}-2ca+{{a}^{2}})=0$

$\Leftrightarrow $${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0$

Vì ${{(a-b)}^{2}}\ge 0;\,{{(b-c)}^{2}}\ge 0;\,{{(c-a)}^{2}}\ge 0$ với mọi $a,b,c$ nên

${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=0$

$\Leftrightarrow $$\left\{ \begin{align}& a-b=0 \\ & b-c=0 \\ & c-a=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& a=b \\ & b=c \\ & c=a \\ \end{align} \right.$$\Leftrightarrow $$a=b=c$\

Khi đó: $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1$

$M=\left( 1+\frac{a}{b} \right)\left( 1+\frac{b}{c} \right)\left( 1+\frac{c}{a} \right)=\left( 1+1 \right)\left( 1+1 \right)\left( 1+1 \right)=8$

Bài 1: Cho $3\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)={{\left( a+b+c \right)}^{2}}$.

Tính giá trị biểu thức $M=\left( 1+\frac{a}{b} \right)\left( 1+\frac{b}{c} \right)\left( 1+\frac{c}{a} \right)$