[SOLVED] Find a+b+c

[anemone]

$a,\,b$ and $c$ are positive integers that satisfy the inequality $ab+3b+2c>a^2+b^2+c^2+3$. Evaluate $a+b+c$.

 

[kaliprasad]

Spoiler

We have


$a^2 + b^2 + c^2 + 3 < ab + 3b +2c$


or $(a^2-ab) + (b^2 - 3b) + (c^2 - 2c) + 3 < 0$


or $(a-\frac{b}{2})^2 - \frac{b^2}{4} + (b^2 - 3b) + (c^2 - 2c) + 3< 0$


Or $(a-\frac{b}{2})^2 + (3 \frac{b^2}{4} - 3b) + (c^2 - 2c) +3 < 0$


Or $(a-\frac{b}{2})^2 + 3(\frac{b^2}{4} - b + 1) + (c^2 - 2c) < 0$


Or $(a-\frac{b}{2})^2 + 3(\frac{b}{2} -1)^2 + (c^2 - 2c+1) < 1$


Or $(a-\frac{b}{2})^2 + 3(\frac{b}{2} -1)^2 + (c-1)^2 < 1$


For these to be true each of the terms has to be zero.


(it can be seen that each of the terms which is squared if not zero has to be minimum $\frac{1}{4}$ giving sum 1 which is not true)


So b = 2 a = 1 and c = 1 is the solution giving a+b+c = 4