Question

Deduce that if a, b and c are real and
unequal, show that
(a + b + c)² > 3(ab + bc + ac)

Answer 1

$\displaystyle \sf (a+b+c)^2 > 3(ab+bc+ac)\\\\ a^2+b^2+c^2+2(ab+bc+ac) > 3(ab+bc+ac)\\\\ a^2+b^2+c^2 -(ab+bc+ac) > 0 \\\\ a^2+b^2+c^2-ab-bc-ac > 0 \ \ x( 2) \\\\ 2a^2+2b^2+2c^2-2ab-2bc-2ac > 0 \\\\ \underbrace{\sf a^2-2ab+b^2}_{(a-b)^2} + \underbrace{\sf a^2-2ac+c^2}_{(a-c)^2} +\underbrace{\sf b^2-2bc+c^2}_{(b-c)^2} > 0 \\\\\\ \underbrace{\sf (a-b)^2}_{\geq 0 } +\underbrace{\sf (a-c)^2}_{\geq 0 } +\underbrace{\sf (b-c)^2}_{\geq 0 } > 0$

$\displaystyle \sf if \ a,b\ e \ c \ are \ unequal \to\ a\neq b \neq c \ \ so : \\\\ \underbrace{\sf (a-b)^2}_{ > 0 } +\underbrace{\sf (a-c)^2}_{ > 0 } +\underbrace{\sf (b-c)^2}_{ > 0 } > 0 \\\\\\ \huge\boxed{\sf proved}\checkmark$