Ta thấy: [TEX]a^2=b^2+b+\frac{1}{3}=(b+\frac{1}{2})^2+\frac{1}{12} > 0 \Rightarrow a > 0[/TEX]
Tương tự thì [TEX]b,c > 0[/TEX]
Giả sử [tex]a=max\left \{ a;b;c \right \}[/tex] [TEX] \Rightarrow a \geq b \Rightarrow b^2+b+\frac{1}{3} \geq c^2+c+\frac{1}{3} \Rightarrow (b-c)(b+c+1) \geq 0 \Rightarrow b \geq c \Rightarrow c^2+c+\frac{1}{3} \geq a^2+a+\frac{1}{3} \Rightarrow (c-a)(c+a+1) \geq 0 \Rightarrow c \geq a[/TEX]
Mà [tex]a=max\left \{ a;b;c \right \}[/tex] nên [TEX]a=b=c[/TEX]