Is smooth convex function with unique minimizer strictly convex? Here the strictly convexity of a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ means
$$f((1-\lambda)x+\lambda y)<(1-\lambda)f(x)+\lambda f(y)\text{ for all }x\ne y,\lambda\in (0,1).$$
Without the smooth assumption, function $f(x)=|x|$ is an immediate counterexample. However, with the smoothness, I don't know how to proceed.
Edit: in this answer, 'smooth' refers to $C^{\infty}$.
We can find a counterexample in dimension $n=1$ by taking $x \mapsto |x|$ and smoothing it out in a particular way. Take a smooth function $\rho \colon \mathbb{R} \rightarrow [0,1]$ with the following properties: (i) $\rho$ is constantly equal to 1 inside the interval $[-0.5,0.5]$ and constantly equal to 0 outside of the interval $[-1,1]$; (ii) $\rho$ is increasing on the interval $(-\infty,0)$ and decreasing on the interval $(0,\infty)$; (iii) $\rho$ is convex on $(-\infty,-0.75) \cup (0.75,\infty)$ and concave on $(-0.75,0.75)$. For the existence of such a function, see Wikipedia [1].
Now define $f \colon \mathbb{R} \rightarrow \mathbb{R}$ by $$ f(x) = x^2 \rho(x) - \frac{9}{16} \rho(x) + 4 \int_0^{|x|} (1-\rho(y)) dy. $$ Then $f$ is smooth despite the appearance of $|x|$ in the integral because of (i). Also by (i), for $|x| > 1$, $$ f(x) = 4 |x| - 4 \int_0^1 \rho(y) dy, $$ so $f$ is not strictly convex. The derivative of $f$ is $$ f'(x) = 2 x \rho(x) + x^2 \rho'(x) - \frac{9}{16} \rho'(x) + 4 \mathrm{sign}(x) (1-\rho(|x|)). $$ The second derivative of $f$ is $$ f''(x) = 2 \rho(x) + 2 x \rho'(x) + 2 x \rho'(x) + x^2 \rho''(x) - \frac{9}{16} \rho''(x) - 4 \rho'(|x|). $$ Note that $x\rho'(x) = -|x\rho'(x)|$ and $\rho'(|x|) = -|\rho'(x)|$ by (ii). Thus, $$ f''(x) = 2 \rho(x) + \left( x^2 - \frac{9}{16} \right) \rho''(x) + 4 |\rho'(x)| (1 - |x|). $$ The first term $2 \rho(x)$ is always nonnegative as $\rho$ takes values in $[0,1]$. The last term $4 |\rho'(x)| (1 - |x|)$ is always nonnegative by (i). The second term $\left( x^2 - \frac{9}{16} \right) \rho''(x)$ is always nonnegative by (iii). Thus, $f''$ is always nonnegative, which shows that $f$ is convex. Finally, $f$ has a unique minimizer at 0 because it is strictly convex in a neighbourhood of 0, which follows from $f(x) = x^2 - 9/16$ for $|x| < 0.5$.
[1] https://en.wikipedia.org/wiki/Bump_function