2 vế đồng nhất nên ta
chuẩn hóa abc=1
[tex]\left\{\begin{matrix} p=a+b+c & & \\ q=ab+bc+ac & & \\ r=abc & & \end{matrix}\right.[/tex]
[tex]a^2+b^2+c^2+\frac{9}{(a+b+c)}-2(ab+bc+ac)=p^2+\frac{9}{p}-4q[/tex]
Theo bđt schur ta có : [tex]p^3+9r\geq 4pq =>q\leq \frac{p^3+9}{4p}[/tex]
[tex]p^2+\frac{9}{p}-4q\geqslant p^2+\frac{9}{p}-4\frac{p^3+9}{4p}=\frac{4p^3+36-4p^3-36}{4p}=0[/tex]
=>đpcm