If  f(x) = x2+2bx+2c2  and  g(x) =  −x2−2cx+b2  are such that  min f(x) > max g(x),  then the relation between b and c is

f(x)=x2+2bx+2c2=(x+b)2+2c2−b2⇒ minf(x)=2c2−b2 Also  g(x)=−x2−2cx+b2 =b2+c2−(x+c)2 So maxg(x)=b2+c2 Since minf(x)>maxg(x) So 2c2−b2>b2+c2⇒ c2>2b2⇒|c|>2|b|