Let $M_n$ be the vector space of real $n \times n$ matrices.
Define $f:M_n \to \mathbb{R}$ by $f(A)=\text{dist}^2(A,\text{SO}(n))$, where the distance is measured using the Frobenius (Euclidean) norm. It is known that $f$ is not convex.
Question: Does there exist a $c>0$ such that $$ f(tA+(1-t)B)\le c \big(tf(A)+(1-t)f(B)\big)$$ for every $A,B \in M_n, t \in [0,1]$?
If so, can we say something about the optimal $c$?
($c$ certainly cannot be arbitrarily close to $1$, since this would imply $f$ is convex).
If it helps, here is an explicit description of $f(A)$ in terms of the singular values of $A$:
$$ f(A)=(\sigma_1'-1)^2 + \sum_{i=2}^n (\sigma_i-1)^2,$$
where $\sigma_1 \le \sigma_2 \le \dots \le \sigma_n$ are the singular values of $A$, and $\sigma_1'=\text{sign} (\det A) \sigma_1$.
In particular, if $\det A \ge 0$, then
$$ f(A)= \sum_{i=1}^n (\sigma_i-1)^2.$$
As commented by Max, there is no such $c>0$. Suppose $n$ is even, and take $A=Id,B=-Id,t=\frac{1}{2}$. Then,
$$f(tA+(1-t)B)\le c \big(tf(A)+(1-t)f(B)\big)$$ becomes
$$f(0)\le \frac{1}{2}c \big(f(A)+f(B)\big)=0,$$
since $A,B \in \text{SO}(n)$. This implies $n=f(0) \le 0$ which is a contradiction.
When $n$ is odd, we can embed $\text{SO}(n-1)$ inside $\text{SO}(n)$ and use the same argument.