What is $\max{(1/\alpha+1/\beta+|1/\gamma|+|1/\delta|)}$? Consider a polynomial $(\alpha ,\beta >0)$,$f(x)=x^3/\alpha+x^2/\beta+x/\gamma+1/\delta$.
If $|f(x)|\leq 1$ for $|x|\leq 1$ then $\max{(1/\alpha+1/\beta+|1/\gamma|+|1/\delta|)}$ is what?
Intuitively it seems that the value should be within the range $1$ to $10$....just plugged in random values.But I'm not able to solve it....

I think is  $$\max \left\{ \dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{|\gamma|}+\dfrac{1}{|\delta|}\right\}= 7$$
In fact,it is 1996 IMO shortist problem :
Let $P(x)=ax^3+bx^2+cx+d$, where $a,b,c,d$ are real numbers, if $|x|\le 1\implies|P(x)|\le 1$, show that:-
$$|a|+|b|+|c|+|d|\le 7$$
Proof: it is clear
$$|P(1)|=|a+b+c+d|\le 1\space,|P(-1)|=|-a+b-c+d|\le 1$$
$$\left|P\left(\dfrac{1}{2}\right)\right|=\left|\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d\right|\le 1\space,\left|P\left(-\dfrac{1}{2}\right)\right|=\left|-\dfrac{1}{8}a+\dfrac{1}{4}b-\dfrac{1}{2}c+d\right|\le 1$$
Note 
\begin{align*}
&|\lambda\cdot a+b|=\left|\dfrac{4}{3}(\lambda\cdot a+b+\lambda\cdot c+d)-2\left(\dfrac{\lambda}{8}a+\dfrac{1}{4}c+\dfrac{\lambda}{2}c+d\right)+\dfrac{2}{3}\left(-\dfrac{\lambda}{8}a+\dfrac{1}{4}b-\dfrac{\lambda}{2}c+d\right)\right|\\
&\le\dfrac{4}{3}+2+\dfrac{2}{3}=4
\end{align*}
where $\lambda=\pm 1$
so we have
$$|a|+|b|=\max{\{|a+b|,|-a+b|\}}\le 4$$
\begin{align*}
&|\lambda c+d|=\left|-\dfrac{1}{3}\left(\lambda a+b+\lambda c+d\right)+2\left(\dfrac{\lambda}{8}a+\dfrac{1}{4}b+\dfrac{\lambda}{2}c+d\right)-\dfrac{2}{3}\left(-\dfrac{\lambda}{8}a+\dfrac{1}{4}b-\dfrac{\lambda}{2}c+d\right)\right|\\
&\le \dfrac{1}{3}+2+\dfrac{2}{3}=3
\end{align*}
so we have$$|c|+|d|\le\max{\{|c+d|,|-c+d|\}}\le 3$$
then 
$$|a|+|b|+|c|+|d|\le 7$$
