# If ab + bc + ca is equal or greater than 2, what is the minimum possible value for a^2 + b^2 + c^2? This was on a noncalculator paper

Jun 24, 2018

2

#### Explanation:

${\left(a - b\right)}^{2} + {\left(b - c\right)}^{2} + {\left(c - a\right)}^{2}$
$q \quad = 2 \left({a}^{2} + {b}^{2} + {c}^{2}\right) - 2 \left(a b + b c + c a\right)$

This shows that

$2 \left({a}^{2} + {b}^{2} + {c}^{2}\right) - 2 \left(a b + b c + c a\right) \ge 0$
(since a sum of squares of real numbers can not be negative)

Thus
${a}^{2} + {b}^{2} + {c}^{2} \ge a b + b c + c a \ge 2$

Thus the minimum possible value for ${a}^{2} + {b}^{2} + {c}^{2}$ is $2$ (this minimum can be achieved if $a = b = c = \sqrt{\frac{2}{3}}$)