1/ Cho a,b,c $\geq$0 ,t/m:$a^{2} +4b^{2}+9c^{2}=14
CMR: 3b+8c+abc\leq 12$
Lời giải 2:Viết lại BĐT cần chứng minh:
\[6b + 16c + 2abc \le 24\]
Ta có:
\[\begin{array}{l}
VT \le 3\left( {{b^2} + 1} \right) + 8\left( {{c^2} + 1} \right) + 2abc = 3{b^2} + 8{c^2} + 2abc + 11 \\
= {a^2} + 4{b^2} + 9{c^2} - \left( {{a^2} + {b^2} + {c^2}} \right) + 2abc + 11 = 2abc - \left( {{a^2} + {b^2} + {c^2}} \right) + 25 \\
\end{array}\]
Do đó, ta chỉ cần chứng minh:
\[\begin{array}{l}
2abc - \left( {{a^2} + {b^2} + {c^2}} \right) \le - 1 \Leftrightarrow {a^2} + {b^2} + {c^2} \ge 1 + 2abc \\
\Leftrightarrow 14\left( {{a^2} + {b^2} + {c^2}} \right) \ge 14 + 28abc \\
\Leftrightarrow \left( {{a^2} + 4{b^2} + 9{c^2}} \right) + 13{a^2} + 10{b^2} + 5{c^2} \ge 14 + 28abc \\
\Leftrightarrow 13{a^2} + 10{b^2} + 5{c^2} \ge 28abc \\
\Leftrightarrow \left( {13{a^2} + 10{b^2} + 5{c^2}} \right)\sqrt {{a^2} + 4{b^2} + 9{c^2}} \ge 28\sqrt {14} abc\left( * \right) \\
\end{array}\]
Mà theo BĐT AM-GM, ta có:
\[\begin{array}{l}
13{a^2} + 10{b^2} + 5{c^2} \ge 28\sqrt[{28}]{{{a^{26}}{b^{20}}{c^{10}}}} \\
{a^2} + 4{b^2} + 9{c^2} \ge 14\sqrt[{14}]{{{a^2}{b^8}{c^{18}}}} \\
\Rightarrow \left( {13{a^2} + 10{b^2} + 5{c^2}} \right)\sqrt {{a^2} + 4{b^2} + 9{c^2}} \ge 28\sqrt {14} \sqrt[{28}]{{{a^{26}}{b^{20}}{c^{10}}}}.\sqrt {\sqrt[{14}]{{{a^2}{b^8}{c^{18}}}}} = 28\sqrt {14} abc \\
\end{array}\]
Nên (* ) đúng hay ta có đpcm.