Finding the maximal $t$ satisfying a family of inequalities Given $c \in (0,1)$, find the maximal positive $t$ satisfying the following:
$$\forall n \in \{1,2,\ldots \}: 1+\frac{c}{n+(1-c)} \le \left(1+\frac{1}{n+t}\right)^{c}$$
My progress thus far:


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*A special case is $c = \frac{1}{2}$: the inequality simplifies to $3t-1 \le n(1-4t)$, which implies $t \le \frac{1}{4}$. $t=\frac{1}{4}$ actually works.

*The inequality $\left(1+\frac{1}{n+t}\right)^{c} \le \frac{c}{n+t}$ shows $t \le 1-c$.

The correct $t$ is $\frac{1-c}{2}$, as I was able to guess in the comments using vadim123's calculations.
Proof: define $f(x)=\ln(1+\frac{1}{\frac{1}{x}+t})^c - \ln(1 + \frac{c}{\frac{1}{x}+1-c})$. We are given that $f(\frac{1}{n}) \ge 0$ for any $n \ge 1$.
Simplifying $f$, we find $f(x)=c \ln(1+\frac{x}{1+tx})-\ln(1+\frac{cx}{1+x(1-c)})$. 
The first derivative is $f'(x)=c (\frac{1}{(1+tx)(1+(t+1)x)}-\frac{1}{(1+x)(1+(1-c)x)})$
If $t=\frac{1-c}{2}$, the first derivative is positive for positive $x$, since it amounts to $(1+x(1-c))(1+x) \ge (1+x\frac{1-c}{2})(1+x\frac{3-c}{2})$, which upon expanding it becomes $x^2(1-c^2) \ge 0$. This shows that $f$ is increasing, hence $f(\frac{1}{n}) \ge f(0) = 0$.
In general, $f'(x) \ge 0$ iff $x(1-c-t(t+1))\ge 2t+c-1$ (assuming $x \ge 0$). Assume $1-c\ge t > \frac{1-c}{2}$ also works. Since $ 2t+c-1 > 0$, $f$ is necessarily decreasing in a small right neighborhood of $0$, so $f(\frac{1}{n})$ is negative for big enough $n$ since $f(0)=0$.
