If the roots of the equation ax2+bx+a21+b21+c21āa1b1āa1c1āb1c1=0 are non real then
2(a−b)+∑(a1−b1)2>0
Let f(x)=ax2+bx+a21+b21+c21−a1b1−a1c1−b1c1
Given non - real roots ⇒ f(x) will have same sign ∀x.
∴ f(0)=a21+b21+c21−a1b1−b1c1−c1a1=12(2a21+2b21+2c21−2a1b1−2b1c1−2c1a1)=12[(a1−b1)2+(b1−c1)2+(c1−a1)2]>0∴ f(−1)>0 ⇒a−b+a21+b21+c21−a1b1−a1c1−b1c1>0⇒2(a−b)+(a1−b1)2+(b1−c1)2+(c1−a1)22>0