<s>[INST] proving an inequality related to $AM\ge GM$ $$a^2+ab+b^2\ge 3(a+b-1)$$
$a,b$ are real numbers
using $AM\ge GM$
I proved that 
$$a^2+b^2+ab\ge 3ab$$
$$(a^2+b^2+ab)/3\ge 3ab$$
how do I prove that $$3ab\ge 3(a+b-1)$$
if I'm able to prove the above inequality then i'll be done
 [/INST] This type of inequality can be proved in the similar way like this:
$$ a^2 + ab + b^2 - 3(a + b - 1) = a^2 + ab + b^2 - 3a - 3b + 3 $$
$$ = a^2 - 2a + 1 + b^2 - 2b + 1 + ab -a -b + 1 = (a - 1)^2 + (b-1)^2 +(a-1)(b-1) $$ $$\ge 2|a-1||b-1| - (a -1)(b-1) \ge 0$$
The last one comes from AM-GM inequality.
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