# If a,b,c are real numbers prove that (a+b-c)^2+(b+c-a)^2+(c+a-b)^2)>= ab+bc+ca ?

Aug 10, 2016

See explanation...

#### Explanation:

If $a , b , c$ are Real then:

$\left\{\begin{matrix}{\left(a - b\right)}^{2} \ge 0 \\ {\left(b - c\right)}^{2} \ge 0 \\ {\left(c - a\right)}^{2} \ge 0\end{matrix}\right.$

So:

$0 \le \frac{3}{2} \left({\left(a - b\right)}^{2} + {\left(b - c\right)}^{2} + {\left(c - a\right)}^{2}\right)$

$= \frac{3}{2} \left({a}^{2} - 2 a b + {b}^{2} + {b}^{2} - 2 b c + {c}^{2} + {c}^{2} - 2 a c + {a}^{2}\right)$

$= \frac{3}{2} \left(2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2 a b - 2 b c - 2 c a\right)$

$= 3 {a}^{2} + 3 {b}^{2} + 3 {c}^{2} - 3 a b - 3 b c - 3 c a$

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

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

Add $a b + b c + c a$ to both ends and transpose to find:

${\left(a + b - c\right)}^{2} + {\left(b + c - a\right)}^{2} + {\left(c + a - b\right)}^{2} \ge a b + b c + c a$

Aug 10, 2016

See below

#### Explanation:

Given two vectors

$\vec{u} = \left\{a , b , c\right\}$
$\vec{v} = \left\{b , c , a\right\}$

we have

$< \vec{u} , \vec{u} > = {\left\lVert \vec{u} \right\rVert}^{2} \ge < \vec{u} , \vec{v} >$

or equivalently

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

or

$3 \left({a}^{2} + {b}^{2} + {c}^{2}\right) \ge 3 \left(a b + b c + a c\right)$

or

$3 \left({a}^{2} + {b}^{2} + {c}^{2}\right) - 2 \left(a b + b c + a c\right) \ge a b + b c + a c$

but

${\left(a + b - c\right)}^{2} + {\left(b + c - a\right)}^{2} + {\left(c + a - b\right)}^{2} = 3 \left({a}^{2} + {b}^{2} + {c}^{2}\right) - 2 \left(a b + b c + a c\right)$

so concluding

${\left(a + b - c\right)}^{2} + {\left(b + c - a\right)}^{2} + {\left(c + a - b\right)}^{2} \ge a b + b c + a c$