JOMO 5, Short 4

According to Cauchy In-equality: $abc=ab+bc+ca \ge 3(abc)^{\frac{2}{3}} \Rightarrow abc \ge 27$ Sign '=' happens when a=b=c=3. So: $MIN(a+b+c)=\boxed{9}$

The solution is quite simple take abc common out of the equation and since a,b,c>0 This implies abc>0 Therefore the equation now becomes "\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =1." Now for \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } to be equal to one a,b,c can have a minimum of value of 3 only since they are integers. Therefore our answer becomes 3+3+3 which is equal to 9.

Do a simple hit and trial. Starting with 1 and keep a=b=c you will hit the jackpot at 3. :) (a=b=c=3) Easy problem!! Nothing that qualifies this as a level 3 stuff :) IMO

ab+bc+ac=abc equally its same with

1/a+1/b+1/c=1

(a+b+c)/3≥ 3/(1/a+1/b+1/c)

(a+b+c) (1/a+1/b+1/c)≥9

Where 1/a+1/b+1/c=1

So the minimum of the sum a+b+c=9 Please follow me to give you the others challange