MHB Prove Inequality $(a,b,c\geq1)$

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$given :\,\,$ $a,b,c\geq1$

$prove:$

$(1+a)(1+b)(1+c)\geq 2(1+a+b+c)$
 
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Here is my solution.

$$(1+a)(1+b)(1+c) = 1 + a + b + c + ab + bc + ca + abc $$

and since $a,b,c \ge 1$, then $ab + bc + ca \ge a + b + c $ and $abc \ge 1$; the result now follows.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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