Find all values of $c$ for which $(1+c)^k - c^k - kc^{k-1} \ge 0$ For an integer $ k \ge 1$ and and a real number $c$, consider the polynomial function $f_{k,c}:\mathbb R \to \mathbb R$ by $f_{k,c}(z) = (z+c)^k$. Also define $r_{k,c} := f_{k,c}(1) - f_{k,c}(0) - f'_{k,c}(0)=(1+c)^k - c^k - kc^{k-1}$. I'm interested in the set of values of $c$ for which $r_{k,c} \ge 0$, i.e the set $S_k := \{c \in \mathbb R \mid r_{k,c} \ge 0\}$.
The case of even $k$
It is clear that if $k$ is even, then $f_{k,c}$ is convex and so $S_k = \mathbb R = (-\infty,\infty)$.
The case of odd $k$
Experiments seem to show that when $k$ is odd, $S_k$ is always an interval of the form $[c_k,\infty)$, for some $c_k \le 0$ with $r_{k,c_k} = 0$.

Question 1. Is it really true that for odd $k$, one has $S_k = [c_k,\infty)$ for some $c_k  \in (-\infty,0]$ with $r_{k,c_k} = 0$ ?

In case the answer to the above question is affirmative,

Question 2. For large $k$, what is a good estimate for $c_k$ ?

 A: Consider the polynomials
$$
 p_k(x) = (1+x)^k - x^k - k x^{k-1} \, .
$$
The case $p_1(x) = 0$ is not interesting, therefore we will consider $k \ge 2$ only.
$p_k$ is a polynomial with the exact degree $k-2$, and
$$
 p_k'(x) = k p_{k-1}(x) \, .
$$
This property can be used to prove that

*

*If $k\ge 2$ is even then $p_k(x) > 0$ for all $x \in \Bbb R$.

*If $k \ge 3$ is odd then $p_k$ is strictly increasing and has exactly one zero $c_k$. Also $-1 < c_k < 0$.

Sketch of a proof via induction:

*

*The claim surely holds for $p_2(x) = 1$ and $p_3(x) = 3x+1$.

*If $k$ is even and $p_k(x) > 0$ for all $x$ then $p_{k+1}$ is strictly increasing. Also $p_k(-1) = 1-k < 0$ and $p_k(0) = 1 > 0$. It follows that $p_{k+1}$ has exactly one zero $c_k$, and that is in the interval $(-1, 0)$.

*If $k$ is odd and $p_k$ is strictly increasing with exactly one zero $c_k$ then $p_{k+1}$ is decreasing on $(-\infty, c_k]$ and increasing on $[c_k, \infty)$. Also
$$
 p_{k+1}(c_k) = (1+c_k)^{k+1} - c_k^{k+1}-(k+1)c_k^k \\
 = (1+c_k)(c_k^k + kc_k^{k-1}) - c_k^{k+1}-(k+1)c_k^k \\
 = kc_k^{k-1} > 0
$$
shows that the minimum of $p_{k+1}$ is strictly positive.

Therefore, in your notation, $S_k = \Bbb R$ if $k$ is even, and $S_k = [c_k, \infty)$ with some $c_k \in(-1, 0)$ if $k$ is odd.
The following estimate holds for odd $k$:
$$
 -\frac 12 < c_k < d_k := - \frac 12 \cdot \frac{1}{(k2)^{1/(k-1)}} \, ,
$$
so that in particular $\lim_{n \to \infty} c_{2n+1} = -1/2$.
The lower bound follows from $g(-1/2) < 0$, and the upper bound follows from
$$
 g_k(x) > \frac{1}{2^k} - kx^{k-1}
$$
for $x > -1/2$ and determining $d_k$ as the solution of $\frac{1}{2^k} - kx^{k-1} = 0$.
