Ta có : [tex]\frac{a^2}{b-1}+4(b-1)\geq 2\sqrt{\frac{4a^2(b-1)}{(b-1)}}=4a[/tex]
Tương tự ta cũng cs [tex]\left\{\begin{matrix} \frac{b^2}{c-1}+4(c-1)\geq 4a & & \\ \frac{c^2}{a-1}+4(a-1)\geq 4c & & \end{matrix}\right.[/tex]
[tex]=>\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}+4(a+b+c)-12\geq 4(a+b+c)[/tex]
[tex]<=>\frac{a^2}{b-1}+\frac{b^2}{c-1}+\frac{c^2}{a-1}\geq 12[/tex]