Lời giải:
Ta có:
\(\sum \frac{1}{a+ab}\geq \frac{3}{abc+1}\Leftrightarrow \sum \frac{abc+1}{a(b+1)}\geq 3\)
\(\Leftrightarrow \sum \frac{bc}{b+1}+\sum\frac{1}{a(b+1)}\geq 3\)
\(\Leftrightarrow \sum \frac{b(c+1)}{b+1}+\sum \frac{a+1}{a(b+1)}\geq 6\)
BĐT trên luôn đúng vì theo BĐT AM-GM thì:
\(\sum \frac{b(c+1)}{b+1}+\sum \frac{a+1}{a(b+1)}=\frac{b(c+1)}{b+1}+\frac{c(a+1)}{c+1}+\frac{a(b+1)}{a+1}+\frac{a+1}{a(b+1)}+\frac{b+1}{b(c+1)}+\frac{c+1}{c(a+1)}\)
\(\geq 6\sqrt[6]{\frac{abc(a+1)^2(b+1)^2(c+1)^2}{abc(a+1)^2(b+1)^2(c+1)^2}}=6\)
Do đó ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c=1\)
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