Gọi ý giải bài 1.10 + 1.12
Bài 1.10:
\[\begin{array}{l}
a,b,c \ge 0.\frac{a}{{1 + a}} + \frac{{2b}}{{1 + b}} + \frac{{3c}}{{1 + c}} \le 1,CMR:a{b^2}{c^3} \le \frac{1}{{{5^6}}}\\
+ )\frac{a}{{1 + a}} + \frac{{2b}}{{1 + b}} + \frac{{3c}}{{1 + c}} \le 1 \leftrightarrow \frac{1}{{1 + a}} + \frac{2}{{1 + b}} + \frac{3}{{1 + c}} \ge 5\\
= > \frac{1}{{a + 1}} \ge 2 - \frac{2}{{1 + b}} + 3 - \frac{3}{{1 + b}} = \frac{{2b}}{{1 + b}} + \frac{{3c}}{{1 + c}} \ge 5\sqrt[5]{{\frac{{{b^2}}}{{{{\left( {1 + b} \right)}^2}}}.\frac{{{c^3}}}{{{{\left( {1 + c} \right)}^3}}}}}\\
\frac{1}{{1 + b}} \ge \frac{a}{{1 + a}} + \frac{b}{{1 + b}} + \frac{{3c}}{{1 + c}} \ge 5\sqrt[5]{{\frac{a}{{1 + a}}.\frac{b}{{1 + b}}.\frac{{{c^3}}}{{{{\left( {1 + c} \right)}^3}}}}}\\
\frac{1}{{1 + c}} \ge \frac{a}{{1 + a}} + \frac{{2b}}{{1 + b}} + \frac{{2c}}{{1 + c}} \ge 5\sqrt[5]{{\frac{a}{{1 + a}}.\frac{{{b^2}}}{{{{\left( {1 + b} \right)}^2}}}.\frac{{{c^2}}}{{{{\left( {1 + c} \right)}^2}}}}}
\end{array}\]
\[\begin{array}{l}
= > \frac{1}{{1 + a}}.{\left( {\frac{1}{{1 + b}}} \right)^2}.{\left( {\frac{1}{{1 + c}}} \right)^3} \ge {5^6}\frac{{a{b^2}{c^3}}}{{\left( {1 + a} \right){{\left( {1 + b} \right)}^2}{{\left( {1 + c} \right)}^3}}}\\
= > dpcm
\end{array}\]