$\def\R{\mathbb R}f\colon\R^n\to\R$ is a coercive function. Given that $A$ is a positive definite matrix in $\R^{n\times n}$ and $b$ is a vector in $\R^n$,prove the function $g\colon \R^n\to\R$ defined by $g(x)=f(Ax+b)$ is coercive ($x\in \R^n$ in this question).
You may use the fact that $x^\top Ax≥λ_\min (A) x^\top x$ for any nonzero $x\in\R^n$ where $λ_\min (A)$ is the smallest eigenvalue of $A$. ($x^\top$ means transpose of $x$).
Even though there is a hint, I still could not get it through. I was thinking that if I can prove that \|Ax+b\|^2 > P(λ_\min (A)) \cdot \|x\|^2 = P(λ_\min (A)) \cdot x^\top x > P(λ_\min (A)) \cdot r_1^2$$ where $P$ is a positive polynomial, $r_1$ is found by the fact that $f$ is coercive: for any $M>0$, there exist $r_1>0$ such that for any $\|x\|\ge r_1$, $f(x)>M$. If I can construct such $P$, then let $P(λ_\min (A))r_1^2 = r^2$, then for any $M>0$, there exists $r>0$, such that for any $\|x\|>r$, $\|Ax+b\|>r_1$, thus $g(x)=f(Ax+b)>M$. Thus $g$ is coercive. But I just could not find a working $P$.