Prove that $|P(x)| ≤|4x^3-3x|$ for all $|x| >1$

Question

$$P(x)=ax^3+bx^2+cx+d; |P(x)| <1 \text{ for all } |x| <1$$ Prove that $$|P(x)| ≤|4x^3-3x|\text{ for all } |x| >1$$

Answer 1

Given $|P(x)|<1\ \forall\ x<1\ \ \ \ \ $ -----(I)

If $a\ne0$ then $\displaystyle\lim_{x\to-\infty}|P(x)|=\infty$, this violates (I), hence $a=0$.

Similarly for $b$ and $c$,

if $b\ne0$ then $\displaystyle\lim_{x\to-\infty}|P(x)|=\infty$, this violates (I), hence $b=0$,

if $c\ne0$ then $\displaystyle\lim_{x\to-\infty}|P(x)|=\infty$, this violates (I), hence $c=0$

The above implies that $-1<d<1$ and $P(x)=d$.

Also $4x^3-3x>1\ \forall\ x>1.$

Hence, $$|P(x)| ≤|4x^3-3x|\ \forall\ x >1$$