(a+b-c)^2 +(a+c-b)^2 + (b+c-a)^2= 3a^2 + 3b^2 + 3c^2 - 2ab-2bc-2ca -----------1st equation
ab+bc+ca---------------2nd equation
adding 2ab+2ca+2bc to both equations we get
3a^2+3b^2+3c^2------------------- 1st equation
3ab +3bc+3ca---------------2nd equation
a^2 +b^2+c^2 -ab-ac-bc = 0.5 { (a-b)^2 + (b-c)^2 + (c-a)^2} >0 for distinct a,b,c Hence a^2 + b^2 +c^2 > ab+bc+ac
As per question
3a^2 +3b^2+3c^2 > 3ab +3ac + 3bc
(a-b+c)^2 +(b+c-a)^2 + (c+a-b)^2 > ab +bc+ca