When $[a,b]=[-1,1]$, the $\inf$ on the right-hand side is attained (uniquely) by the monic Chebyshev polynomial $T_{n+1}$. It is well known that its roots belong to $[-1,1]$ and are simple. 

For a general interval $[a,b]$, using the linear map 
$$f(x)=\left(\frac{2}{b-a}\right)x-\left(\frac{b+a}{b-a}\right)$$
that sends $[a,b]$ onto $[-1,1]$, it is clear that the $\inf$ on the right-hand side is attained by
$$
\left(\frac{b-a}{2}\right)^{n+1} T_{n+1}(f(x)),$$
which, of course, has all its (simple) roots in $[a,b]$.

Actually, a more general result holds true. Replace $[a,b]$ with any compact subset $K$ of the complex plane $\mathbb{C}$. Set, for a polynomial $P$,
$$\|P\|_{K}=\sup_{z\in K}|P(x)|,$$
and let $T_{n}$ be a polynomial that achieves the minimum of $\|P\|_{K}$ among all monic polynomials of degree $n$.

**Claim**: All the zeros of $T_{n}$ belong to the convex hull of $K$.

Indeed, assume that $z_{0}$ is a root of $T_{n}$ that does not belong to the convex hull of $K$. Then, $K$ lies in a cone with vertex at $z_{0}$, of opening $<\pi$. Choose a $z_{1}$ on the bisector $L$ of that cone, sufficiently close to $z_{0}$ so that $K$ lies in the half-plane, delimited by the perpendicular to $L$ at $z_{1}$ (and not containing $z_{0}$). Since
$$|z-z_{1}|<|z-z_{0}|,\quad z\in K,$$
one gets $\|\tilde T_{n}\|_{K}<\|T_{n}\|_{K}$ where 
$\tilde T_{n}(z)=T_{n}(z)(z-z_{1})/(z-z_{0})$, which contradicts the assumption on $T_{n}$.