$\begin{align} & \text{if }a,b\text{ are real and unequal show that} \\ & \text{ }{{a}^{2}}+{{b}^{2}}>2ab \\ & \text{Deduce that }a,b\text{ and }c\text{ are real and unequal, then} \\ & {{(a+b+c)}^{2}}>3(ab+bc+ca) \\\end{align}$
$\begin{align} & \text{Let }a>b \\ & a-b>0 \\ & \therefore {{(a-b)}^{2}}>0 \\ & {{a}^{2}}-2ab+{{b}^{2}}>0 \\ & {{a}^{2}}+{{b}^{2}}>2ab----(i)\text{ }(QED) \\ & {{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ac)--(ii) \\ & {{a}^{2}}+{{b}^{2}}>2ab----(iii) \\ & {{a}^{2}}+{{c}^{2}}>2ac----(iv) \\ & {{b}^{2}}+{{c}^{2}}>2bc----(v) \\ & \text{Adding (ii),(iii),(iv)} \\ & 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}>2ab+2ac+2bc \\ & \therefore {{a}^{2}}+{{b}^{2}}+{{c}^{2}}>ab+ac+bc---(vi) \\ & \text{Substituting (vi) into (ii) we have} \\ & {{(a+b+c)}^{2}}>ab+ac+bc+2(ab+bc+ac) \\ & {{(a+b+c)}^{2}}>3(ab+bc+ac) \\ & \\\end{align}$