Đề bài: Cho $a,b,c$ là các số thực dương thỏa mãn điều kiện $ab+bc+ca=1$. Chứng minh rằng: \[\frac{{{a}^{3}}}{1+9{{b}^{2}}ac}+\frac{{{b}^{3}}}{1+9{{c}^{2}}ab}+\frac{{{c}^{3}}}{1+9{{a}^{2}}bc}\ge \frac{{{\left( a+b+c \right)}^{3}}}{18}\]

Lời giải

Ta có:

\[\frac{{{a}^{3}}}{1+9{{b}^{2}}ac}+\frac{{{b}^{3}}}{1+9{{c}^{2}}ab}+\frac{{{c}^{3}}}{1+9{{a}^{2}}bc}=\frac{{{a}^{4}}}{a+9{{a}^{2}}{{b}^{2}}c}+\frac{{{b}^{4}}}{b+9a{{b}^{2}}{{c}^{2}}}+\frac{{{c}^{4}}}{c+9{{a}^{2}}b{{c}^{2}}}\]

\[\ge \frac{{{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}^{2}}}{a+b+c+9{{a}^{2}}{{b}^{2}}c+9a{{b}^{2}}{{c}^{2}}+9{{a}^{2}}b{{c}^{2}}}=\frac{{{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}^{2}}}{a+b+c+9abc\left( ab+bc+ca \right)}\]

Mặt khác:
 \[\left\{ \begin{array}{l}
9abc = 3\sqrt[3]{{abc}}.3\sqrt[3]{{{{\left( {abc} \right)}^2}}} \le \left( {a + b + c} \right)\left( {ab + bc + ca} \right) = a + b + c\\
{\left( {{a^2} + {b^2} + {c^2}} \right)^2} \ge \frac{{{{\left( {a + b + c} \right)}^4}}}{9}
\end{array} \right.\]

Khi đó:
  \[\begin{array}{*{20}{c}}
{\frac{{{a^4}}}{{a + 9{a^2}{b^2}c}} + \frac{{{b^4}}}{{b + 9a{b^2}{c^2}}} + \frac{{{c^4}}}{{c + 9{a^2}b{c^2}}} \ge \frac{{\frac{{{{\left( {a + b + c} \right)}^4}}}{9}}}{{2\left( {a + b + c} \right)}} = \frac{{{{\left( {a + b + c} \right)}^3}}}{{18}}}&{\left( {dpcm} \right)}
\end{array}\]