- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,802

- Mar 31, 2013

- 1,331

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

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

Last edited by a moderator: