$f(t)=(\text{det}(A+tB))^{\frac{1}{d}}$ is concave if $f(t)\leq f(0)+tf'(0)$? Let $A$, $B$ be $d$ by $d$ matrices where $A$, $A+B$ are symmetric positive semi-definite matrices and $\text{det}B=0$. $f(t)=(\text{det}(A+tB))^{\frac{1}{d}}$
is my function where $t\in[0,1]$.

If $f(t)\leq f(0)+tf'(0)$, then $f$ is concave over $[0,1]$.

I just asked To be concave on $[0,1]$, $f(t)\leq f(0)+tf'(0)$ is enough?
for general function, but turns out there is a counterexample. 
$f$ is special here, which is a composition of exponent function
and determinent, it could be true. Thanks in advance.
My trial : $f(t)=(\text{det}(A+tB))^{\frac{1}{d}}$
looks like a exponent $\frac{1}{d}$ of a polynomial function of order
$d$. then $f(t)=(a_{1}t^{d}+\cdots)^{\frac{1}{d}}$. 

Second trial : Let $f(t)=(a_{d}t^{d}+\cdots+a_{0})^{1/d}$.
$tf(1/t)=(\det(tA+B))^{1/d}$, LHS$=(a_{d}+\cdots+a_{0}t^{d})^{1/d}$
take $t=0$, then $a_{d}=\text{det}B=0$.
So $f(t)=(a_{d-1}t^{d-1}+\cdots+a_{0})^{1/d}$
We can see $f$ is smooth. Let's claim $f$ doesn't have zero on $(0,1)$.
Suppose not, i.e. $f(z)=0$ for some $z\in(0,1)$. We know $f(1)=\left(\text{det}(A+B)\right)^{1/d}\geq0$.
Then, we can change $A$ into $A+zB$ then the condition $\tilde{f}(t)\leq\tilde{f}(0)+t\tilde{f}'(0)$
doesn't make sence. So $f$
doesn't have any zero on $(0,1)$.
Here, I can't go farther.
 A: Some headway... Let $C(t) = A+tB$ for $t\in[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $\det C(t) \ge 0$ and thus $f(t)$ is well defined for every $t\in[0,1]$. 
One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= \det C(t)$  for notational brevity. Then we have
$$ f'(t) = \frac{d}{dt} (g(t)^s ) =sg(t)^{s-1} g'(t)$$
and
$$ f''(t) = s(s-1) g(t)^{s-2} g'(t)^2 + sg(t)^{s-1} g''(t)$$
Now observe that


*

*$\frac{d}{dt} [detA(t)]=detA(t) \operatorname{tr}[A^{-1}(t)\, \frac{d}{dt} A(t)]$

*$\frac{d^2}{dt^2} det(A(t)) = \det \left(A\right) \left( \left(\operatorname{tr}\left(A^{-1} A_t\right)\right)^2 + \operatorname{tr}\left(A^{-1} A_{tt}\right)-\operatorname{tr}\left(A^{-1}A_t A^{-1} A_t\right)\right)$
so
\begin{align}f''(t) 
&= s(s-1)g(t)^s [\operatorname{tr}(C^{-1} B)]^2 + s g(t)^s\left([\operatorname{tr}(C^{-1} B)]^2 
   +0-\operatorname{tr}(C^{-1}BC^{-1}B) \right) \\
&= sg(t)^s \left( s [\operatorname{tr}(C^{-1} B)]^2 
   -\operatorname{tr}((C^{-1}B)^2) \right) \end{align}
Similarly to the linked question, it seems that the given inequality $ f(t) \le f(0) + tf'(0)$, 
i.e.,
$$g(t)^s \le g(0)^s + tsg(0)^{s}\operatorname{tr}(A^{-1}B)$$
is not helpful. It seems like we need to know an inequality between traces of $D:=C^{-1}B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that
$$ s(\operatorname{tr} D)^2 - \operatorname{tr}(D^2) = s(\sum_i \lambda_i )^2 - (\sum_i \lambda_i^2) $$
If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
$$ \det M = \frac12((\operatorname{tr}M)^2 - \operatorname{tr}(M^2) ) $$
so we can notice that 
$$ f''(t) = \frac12g(t)^{1/2} \left( -\frac12\operatorname{tr}((C^{-1}B)^2)\right) < 0$$
which proves the result true in this special case, without using the inequality $f(t) \le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.
