For real values of x, the expression (x−b)(x−c)(x−a) will assume all real values provided

Let m=(x−b)x−a(x−c)  ⇒x2−(b+c+m)x+(bc+am)=0Since x is real, we must have  (b+c+m)2−4(bc+am)≥0⇒m2+2(b+c−2a)m+(b−c)2≥0 for all m⇒ 4(b+c−2a)2−4(b−c)2≤0⇒ (b+c−2a)2−(b−c)2≤0⇒ (b+c−2a+b−c)(b+c−2a−b+c)≤0⇒ 2(b−a)2(c−a)≤0⇒ (a−b)(a−c)≤0⇒ b≤a≤c or, c≤a≤b