Question

If a+b+c=0 then prove that, [(b+c)²/3bc]+[(c+a)²/3ac]+[(a+b)²/3ac]

Answer 1

(b+c)²/3bc+(c+a)²/3ac+(a+b)²/3ab
=(b²+2bc+c²)/3bc+(c²+2ac+a²)/3ac+(a²+2ab+b²)/3ab
=(ab²+2abc+ac²+bc²+2abc+a²b+a²c+2abc+b²c)/3abc
={ab(a+b)+bc(b+c)+ac(a+c)+6abc}/3abc
=(-abc-abc-abc+6abc)/3abc [∵, a+b+c=0,∴,a+b=-c,b+c=-a,a+c=-b]
=(6abc-3abc)/3abc
=3abc/3abc
=1 (Proved)