I am just starting to learn these inequality problems, I can show some things with AM>=GM, can someone give it a shot on solving fully!

NICKOMODE

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May 8, 2021
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Given that
$(a+1)(b+1)(c+1) = 8$
, and
$a, b, c \ge 0$
show that
$abc \le 1$
.
 
Given that
$(a+1)(b+1)(c+1) = 8$
, and
$a, b, c \ge 0$
show that
$abc \le 1$
.


We are not here to do problems fully for you. Consider looking at the contrapositive:

If abc > 1, then it is either the case that 1) \(\displaystyle \ (a + 1)(b + 1)(c + 1) \ne 8, \)
or 2) exactly two of a, b, c will be negative, or 3) both conditions will be true together.

Please post some attempts.
 
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