Irska

Define $M(a,b,c)={\max }_{x\in [-1,1]}|x^3+a{x}^2+bx+c|$. Observe that $$\underset{x\in [-1,0]}{\max }|x^3+a{x}^2+bx+c|=\underset{x\in [0,1]}{\max }|-x^3+a{x}^2-bx+c|.$$ Since $\max (|\alpha +\beta |,|\alpha -\beta |)=|\alpha |+|\beta |$, then $$M(a,b,c)=\underset{x\in [0,1]}{\max }(|x^3+bx|+|a{x}^2+c|)\ge \underset{x\in [0,1]}{\max }|x^3+bx|=M(0,b,0).$$ Now we can put $x=\cos \theta$ in $x^3+bx$, where $\theta \in [0,\frac{\pi }{2}]$. Then $x^3+bx={\cos }^3\theta +b\cos \theta =\frac{1}{4}(\cos (3\theta )+3\cos \theta )+b\cos \theta =\frac{1}{4}(\cos (3\theta )+(3+4b)\cos \theta )$. Define $m_{\alpha }={\max }_{\theta \in [0,\frac{\pi }{2}]}|\cos (3\theta )+\alpha \cos \theta |$. Using the values at $\theta =0$ and $\theta =\pi /3$, gives $m_{\alpha }\ge \max (|1+\alpha |,|1-\alpha /2|)\ge 1+|\alpha |/2\ge 1$, all $\alpha$. Clearly $m_0=1$, so ${\min }_{\alpha }{m}_{\alpha }=1$ and this occurs if and only if $\alpha =0$, that is $b=-\frac{3}{4}$. So ${\min }_{b}M(0,b,0)=M(0,-\frac{3}{4},0)=\frac{1}{4}$. So the minimum is $\frac{1}{4}$, which occurs for $x^3-\frac{3}{4}x$.