Question

if a+b+c=0 then prove that a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)=1

Answer 1

a+c+b  = 0

a = -b-c

ca = -bc - c²

2b² + ca = b² - c² + b² - bc

= - (b-c)(a-b)

lly, 2a² + bc = -(a-b)(c-a) and 2c² + ab = -(c-a)(b-c)

Now LHS = -a²/(a-b)(c-a) - b²/(b-c)(a-b) - c²/ (c-a) (b-c)

now by LCM and multiplication

= -{a²b-a²c+b²c-ab²+ac²-bc² / - (a²b-a²c+b²c -ab² + ac² - bc²)]

= 1 = RHS 

Hence Proved.