Ta có: [tex]27a+27a+\frac{1}{a^2}\geq 3.9=27[/tex]
Tương tự cộng vế theo vế ta có: [tex]54(a+b+c)+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq 81\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq 27[/tex]
[tex]\frac{4}{ab+bc+ca}\geq \frac{12}{(a+b+c)^2}=12\Rightarrow ...[/tex]