Maths Question:
$\text{Show that }{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge ab+bc+ca\text{ for all real values of }a,b,c$
Maths Solution:
$\begin{align} & {{(a-b)}^{2}}\ge 0 \\ & {{a}^{2}}+{{b}^{2}}-2ab\ge 0 \\ & {{a}^{2}}+{{b}^{2}}\ge 2ab----(i) \\ & \text{This also hold true for} \\ & {{a}^{2}}+{{c}^{2}}\ge 2ac----(ii) \\ & {{b}^{2}}+{{c}^{2}}\ge 2bc----(iii) \\ & \text{Adding }(i),(ii)\text{ and }(iii)\text{ together} \\ & 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}\ge 2ab+2ac+2bc \\ & \text{Divide through by }2 \\ & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\ge ab+ac+bc \\ \end{align}$
University mathstopic:
