Chứng minh rằng với a,b,c>0 ta có:
[TEX]\frac{1}{4a} + \frac {1}{4b} + \frac{1}{4c} \geq \frac{1}{a + 3b} + \frac{1}{b + 3c} + \frac{1}{c + 3a} \geq \frac{1}{a + 2b + c} + \frac{1}{b + 2c +a} + \frac{1}{c + 2a+ b}[/TEX]
$\begin{array}{l}
\frac{1}{{4a}} + \frac{1}{{4b}} + \frac{1}{{4c}} \ge \frac{1}{{a + 3b}} + \frac{1}{{b + 3c}} + \frac{1}{{c + 3a}} \ge \frac{1}{{a + 2b + c}} + \frac{1}{{b + 2c + a}} + \frac{1}{{c + 2a + b}}\\
ta.co.bdt:\frac{1}{a} + \frac{1}{b} \ge \frac{4}{{a + b}}(tu.chung.{\mathop{\rm mih}\nolimits} )(dau = \leftrightarrow a = b)\\
\to \frac{3}{a} + \frac{1}{c} = \left( {\frac{1}{a} + \frac{1}{a}} \right) + \left( {\frac{1}{a} + \frac{1}{c}} \right) \ge \frac{4}{{2a}} + \frac{4}{{a + c}} \ge \frac{{16}}{{3a + c}}\\
tuong.tu\\
\frac{3}{b} + \frac{1}{a} \ge \frac{{16}}{{3b + a}}\\
\frac{3}{c} + \frac{1}{b} \ge \frac{{16}}{{3c + b}}\\
\to \frac{1}{{4a}} + \frac{1}{{4b}} + \frac{1}{{4c}} \ge \frac{1}{{a + 3b}} + \frac{1}{{b + 3c}} + \frac{1}{{c + 3a}}\\
dau = \leftrightarrow a = b = c\\
ta.co.bdt:\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{{a + b + c}}(dau = \leftrightarrow a = b = c)\\
\to \frac{1}{{a + 3b}} + \frac{1}{{a + 2b + c}} + \frac{1}{{b + 2c + a}} \ge \frac{9}{{3a + 6b + 3a}} = \frac{3}{{a + 2b + c}}\\
tuong.tu\\
\frac{1}{{b + 3c}} + \frac{1}{{a + 2c + b}} + \frac{1}{{b + 2a + c}} \ge \frac{3}{{a + 2c + b}}\\
\frac{1}{{c + 3a}} + \frac{1}{{b + 2a + c}} + \frac{1}{{c + 2b + a}} \ge \frac{3}{{a + b + 2c}}\\
\to \frac{1}{{a + 3b}} + \frac{1}{{b + 3c}} + \frac{1}{{c + 3a}} \ge \frac{1}{{a + 2b + c}} + \frac{1}{{b + 2c + a}} + \frac{1}{{c + 2a + b}}\\
dau = \leftrightarrow a = b = c\\
\to DA
\end{array}$
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