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If $a>0, b>0, c>0$ and $a, b, c$ are distinct, then $(a+$b) $(b+c)(c+a)$ is greater than
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$8 \mathrm{abc}$
$\mathrm{AM}>\mathrm{GM}$
$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}>\sqrt{\mathrm{ab}}, \frac{\mathrm{b}+\mathrm{c}}{2}>\sqrt{\mathrm{bc}}, \frac{\mathrm{c}+\mathrm{a}}{2}>\sqrt{\mathrm{ca}}$
So, $(a+b)(b+c)(c+a)>8 a b c$
$\Rightarrow \frac{\mathrm{a}+\mathrm{b}}{2}>\sqrt{\mathrm{ab}}, \frac{\mathrm{b}+\mathrm{c}}{2}>\sqrt{\mathrm{bc}}, \frac{\mathrm{c}+\mathrm{a}}{2}>\sqrt{\mathrm{ca}}$
So, $(a+b)(b+c)(c+a)>8 a b c$
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