Cho3 số thực dương a, b, c  thỏa  $\left\{\begin{array}{l}ab\ge 12\\ bc\ge 8\end{array}\right.$ .

Chứng minh rằng: $\left(a+b+c\right)+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{8}{abc}\ge \frac{121}{12}$

Áp dụng bất đẳng thức AM - GM ta có:

 $\begin{array}{l}\frac{a}{18}+\frac{b}{24}+\frac{2}{ab}\ge 3\sqrt[3]{\frac{a}{18}.\frac{b}{24}.\frac{2}{ab}}=\frac{1}{2}\\ \frac{a}{9}+\frac{c}{6}+\frac{2}{ca}\ge 3\sqrt[3]{\frac{a}{9}.\frac{c}{6}.\frac{2}{ca}}=1\end{array}$

$\begin{array}{l}\frac{b}{16}+\frac{c}{8}+\frac{2}{bc}\ge 3\sqrt[3]{\frac{b}{16}.\frac{c}{8}.\frac{2}{bc}}=\frac{3}{4}\\ \frac{a}{9}+\frac{c}{6}+\frac{b}{12}+\frac{8}{abc}\ge 4\sqrt[4]{\frac{a}{9}.\frac{c}{6}.\frac{b}{12}.\frac{8}{abc}}=\frac{4}{3}\end{array}$

$\begin{array}{l}\frac{13a}{18}+\frac{13b}{24}\ge 2\sqrt[]{\frac{13a}{18}.\frac{13b}{24}}\ge 2\sqrt[]{\frac{13}{18}.\frac{13}{24}.12}=\frac{13}{3}\\ \frac{13b}{48}+\frac{13c}{24}\ge 2\sqrt[]{\frac{13b}{48}.\frac{13c}{24}}\ge 2\sqrt[]{\frac{13}{48}.\frac{13}{24}.8}=\frac{13}{4}\end{array}$

Cộng theo vế các bất đẳng thức trên ta được:

 $\left(a+b+c\right)+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{8}{abc}\ge \frac{121}{12}$   (đpcm)