MHB Positive Integers: Evaluate $a+b+c$

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The inequality given is \( ab + 3b + 2c > a^2 + b^2 + c^2 + 3 \). By rearranging the terms, it can be analyzed to find relationships between \( a \), \( b \), and \( c \). Testing small positive integer values leads to the conclusion that \( a = 1 \), \( b = 2 \), and \( c = 1 \) satisfy the inequality. This results in \( a + b + c = 4 \). Therefore, the evaluated sum \( a + b + c \) is 4.
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$a,\,b$ and $c$ are positive integers that satisfy the inequality $ab+3b+2c>a^2+b^2+c^2+3$. Evaluate $a+b+c$.
 
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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
 
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