Let $ P( - 1) = y_1,\ P( - 1/2) = y_2,\ P(1/2) = y_3,\ P(1) = y_4$. Then from Lagrange's interpolation formula (or from a linear system of equations) we obtain: $ P(x) = \frac 1 3 (( - 2y_1 + 4y_2 - 4y_3 + 2y_4)x^3 + (2y_1 - 2y_2 - 2y_3 + 2y_4)x^2 +$ $ (\frac 1 2 y_1 - 4y_2 + 4y_3 - \frac 1 2 y_4)x + ( - \frac 1 2y_1 + 2y_2 + 2y_3 - \frac 1 2 y_4)$. It is easy to check, that if $ y_i \in [ - 1,1]$, then $ \pm a \pm b \pm c \pm d \le 7$ for any combination of pluses and minuses. So, $ |a| + |b| + |c| + |d| \le 7$. Remark. This solution applies the technics, which is common for polynomials, bounded on [-1, 1]. Actually we consider Chebyshev's polynomial $ T_n(x)=\cos(n\arccos(x))$ of the same degree and compare it with our polynomial in points, where $ T_n(x)= \pm 1$. In our case it will be a polynomial $ 4x^3-3x$ and the corresponding points are $ -1,\ -1/2,\ 1/2,\ 1.$

Sasha Rybak wrote: It is easy to check, that if $ y_i \in [ - 1,1]$, then $ \pm a \pm b \pm c \pm d \le 7$ for any combination of pluses and minuses. So, $ |a| + |b| + |c| + |d| \le 7$. Can you explain this further? Why is it "easy to check," and how do you check it?

Let $a \le 0, b \le 0$.Then set $Q(x)=-P(x)$ Let $a \le 0,b \ge 0$.Then set $Q(x)=P(-x)$ Let $a \ge 0,b \le 0$.Then set $Q(x)=-P(-x)$ In each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \ge 0,b \ge 0$. $c \ge 0,d \ge 0 \Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \le 1$ $c \ge 0,d < 0 \Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \le 3$ $c < 0,d \ge 0 \Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\frac{4}{3}P(1)-\frac{1}{3}P(-1)-\frac{8}{3}P(\frac{1}{2})+\frac{8}{3}P(-\frac{1}{2}) \le 7$ $c < 0,d < 0 \Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\frac{5}{3}P(1)-4P(\frac{1}{2})+\frac{4}{3}P(-\frac{1}{2}) \le 7$ So we are done!!

This problem is very reminiscent of Chebyshev bounds on polynomials. What I have so far is that $|f(x)| \le |C_3(x)|$ for $|x| \ge 1$, where $C_3(x)$ is the 3rd degree Chebyshev polynomial. However, I cannot seem to isolate into $|a| + |b| + |c| + |d|$? Does anyone have a completion?

orl wrote: Let $ P(x)$ be the real polynomial function, $ P(x) = ax^3 + bx^2 + cx + d.$ Prove that if $ |P(x)| \leq 1$ for all $ x$ such that $ |x| \leq 1,$ then \[ |a| + |b| + |c| + |d| \leq 7.\] Let $f(x)=a+bx+cx^2+dx^3$ and $|f(0)|\leq 1,|f'(0)|\leq 1,|f''(0)|\leq 1,|f(1)|\leq \frac{1}{2}.$ Then for all $x\in [0,1]$ we have $|f(x)|\leq \frac{11}{8}.$

sayantanchakraborty wrote: Let $a \le 0, b \le 0$.Then set $Q(x)=-P(x)$ Let $a \le 0,b \ge 0$.Then set $Q(x)=P(-x)$ Let $a \ge 0,b \le 0$.Then set $Q(x)=-P(-x)$ In each of these cases we see that $Q(x)$ satisfies all the conditions of the problem and the coefficient of $x^3$ and $x^2$ are nonnegative.So wlog let $a \ge 0,b \ge 0$. $c \ge 0,d \ge 0 \Rightarrow |a|+|b|+|c|+|d|=a+b+c+d=P(1) \le 1$ $c \ge 0,d < 0 \Rightarrow |a|+|b|+|c|+|d|=a+b+c-d=P(1)-2P(0) \le 3$ $c < 0,d \ge 0 \Rightarrow |a|+|b|+|c|+|d|=a+b-c+d=\frac{4}{3}P(1)-\frac{1}{3}P(-1)-\frac{8}{3}P(\frac{1}{2})+\frac{8}{3}P(-\frac{1}{2}) \le 7$ $c < 0,d < 0 \Rightarrow |a|+|b|+|c|+|d|=a+b-c-d=\frac{5}{3}P(1)-4P(\frac{1}{2})+\frac{4}{3}P(-\frac{1}{2}) \le 7$ So we are done!! What is Q(x) here

Also if |X| ≤ 1 then isn't X only 1and -1

Let $x_1=-1,x_2=-\frac 12,x_3=\frac 12,x_4=1.$ From Lagrange's interpolation formula$,$ $$\begin{aligned}P(x)&=\sum\limits_{k=1}^4P(x_k)\prod\limits_{j\neq k}\frac {x-x_j}{x_k-x_j}\\&=-\frac 23P(-1)\left(x^3-x^2-\frac x4+\frac 14\right)+\frac 43P\left(-\frac 12\right)\left(x^3-x^2-\frac x4+\frac 14\right)\\&\quad\text{ }-\frac 43P\left(\frac 12\right)\left(x^3+\frac {x^2}2-x-\frac 12\right)+\frac 23P(1)\left(x^3+x^2-\frac x4-\frac 14\right).\end{aligned}$$For $p,q,r,s\in\mathbb R,$ define $(p,q,r,s)=pP(x_1)+qP(x_2)+rP(x_3)+sP(x_4).$ Then $a=\left(-\frac 23,\frac 43,-\frac 43,\frac 23\right),b=\left(\frac 23,-\frac 23,-\frac 23,\frac 23\right),c=\left(\frac 16,-\frac 43,\frac 43,-\frac 16\right),d=\left(-\frac 16,\frac 23,\frac 23,-\frac 16\right).$ Therefore $a+b=\left(0,\frac 23,-2,\frac 43\right),a-b=\left(-\frac 43,2,-\frac 23,0\right)\Rightarrow |a|+|b|=\max\{|a+b|,|a-b|\}\leqslant 4.$ $c+d=\left(0,-\frac 23,2,-\frac 13\right),c-d=\left(\frac 13,-2,\frac 23,0\right)\Rightarrow |c|+|d|=\max\{ |c+d|,|c-d|\}\leqslant 3.$ $\therefore |a| + |b| + |c| + |d|\leqslant 4+3=7.\blacksquare$