Minimum point-wise of convex and scalar 
Let $ f(x) = \log_3 (1+5^x) $ and $ k \geq 0 $. I need to prove that $ \frac{1}{\min \{ f(x), k \}} $ is convex.

My attempt:
Since $ 1+5^x \geq 1$, then $ \log_3 (1+5^x) \geq 0 $. Furthermore, since $ 1+5^x \geq 1$ is monotonically increasing, $ \log_3 (1+5^x) $ all also be so. Thus,
$$ 
\min \{ \log_3 (1+5^x), k \} \rightarrow \min \{ 1+5^x, 3^k \}
$$
So, now I can equivalently try to prove $ \min \{ 1+5^x, 3^k \} $ that looks simpler. I have tried using the convexity definition, but I have not been successful.
 A: Plot for 
$$
g(x,k)=\frac{1}{\min[k,\log_3(1+5^x)]}
$$
with $k = 1$

A: If you rewrite
$$
\frac{1}{\min\{f(x),k\}}=\max\left\{\frac1{f(x)},\frac1k\right\}
$$
then what is left is to prove that
$$
\frac{1}{f(x)}=\frac{1}{\log_3(1+5^x)}=\frac{\ln 3}{\ln(1+5^x)}=\frac{\ln 3}{\ln(1+e^{x\ln 5})}
$$
is convex (as the max of two convex functions is convex). It can be done by proving first that the function
$$
g(t)=\frac{1}{\ln(1+e^t)}
$$
is convex (check that $g''(t)\ge 0$), then 
$$
\frac{1}{f(x)}=\ln 3\cdot g(x\ln 5)
$$
is convex too.
A: I believe you mean $\max$, otherwise the function is not convex.
Suppose that $\max \{f(x), k \} = f(x)$ for $x \geq a$, and $k$ otherwise. The definition of convexity states that the function is convex if every chord lies above the function. It is easy to show that $f(x)$ is convex, by differentiating twice. So every chord on the range $[a,\infty)$ lies above the function. Clearly any chord on $(-\infty, a]$ is simply horizontal and therefore lies under the function. Now consider any chord with endpoints at $x=b, c$ where $b<a<c$. From drawing the graph and noting that $f(x)$ is increasing, it is easy to see that this chord lies above the chord with endpoints $x=a, c$, and this chord in turn lies above the function on that range. Furthermore, this chord clearly lies above $x=k$ over the range $[b,c]$. So combining these observations, the chord lies above the function, and we are done.
A: $\min\{k,\log_3(1+5^x)\}$ is not convex, unless $k=0$.
However, $\max\{k,\log_3(1+5^x)\}$ is, because the maximum of two convex functions is convex. For a proof, see here.
