How can we show $\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $ for all $n \in \mathbb{N}$, $0 \leq c \leq x \leq 1$, $c \neq 1$? 
How can we show $$\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $$ for all $n \in \mathbb{N}$, $0 \leq c \leq x 
 \leq 1, c \neq 1$?

The context is this probability problem, but of course this problem might be of independent interest to inequality enthusiasts.
In my attempt, I have graphed the inequality on Desmos. We might note that the derivative changes sign in the range $c < x < 1$ so it may be unlikely differentiation would be of help.
 A: If $x=c$ or $n=1$ then inequality holds. So consider $x > c, n > 1$:
$$\frac{1-x^n}{1-c^n}+\left(\frac{x-c}{1-c}\right)^n\leq 1\Leftrightarrow
1-x^n+(1-c^n)\left(\frac{x-c}{1-c}\right)^n\leq 1-c^n\Leftrightarrow$$
$$(1-c^n)\left(\frac{x-c}{1-c}\right)^n\leq x^n-c^n\Leftrightarrow
\frac{1-c^n}{(1-c)^n} \leq \frac{x^n-c^n}{(x-c)^n}$$
Consider $f(x)=\frac{x^n-c^n}{(x-c)^n}$, then $$f'(x)=\frac{n x^{n-1}}{(x-c)^n}-\frac{n(x^n-c^n)}{(x-c)^{n+1}}=\frac{nc(c^{n-1}-x^{n-1})}{(x-c)^{n+1}}<0$$
$$f'(x)<0, x\leq 1 \Rightarrow f(1)\leq f(x)\Rightarrow \frac{1-c^n}{(1-c)^n} \leq \frac{x^n-c^n}{(x-c)^n}$$
A: Using the idea proposed by Ivan Kaznacheyeu, we can generalize this to the following inequality:

For any $n \in \mathbb{N}$, suppose $(x_i)_{i=1}^{n} \in [0,1]^n$, $(c_i)_{i=1}^{n} \in [0,1)^n$ with $x_i \geq c_i \ \forall i$. Then $$\frac{1 - x_1x_2 \cdots x_n}{1-c_1c_2 \cdots c_n} + \prod_{i=1}^{n} \left(1 - \frac{1-x_i}{1-c_i}\right) \leq 1$$

Without loss of generality, suppose $x_\ell > c_\ell$ for each index $\ell$. This is fine, since if $x_{\ell} = c_{\ell}$ for some $\ell$, the product $\prod_{i=1}^{n} \left(1 - \frac{1-x_i}{1-c_i}\right)$ is $0$, so the inequality clearly holds. In addition suppose $n \geq 2$. (Otherwise we have equality).
Now, using the same manipulations as that in the answer by Ivan Kaznacheyeu, it suffices to prove $$\frac{1-\prod_{j} c_j}{\prod_{j} (1 - c_j)} \leq \frac{\prod_{j} x_j-\prod_{j} c_j}{\prod_{j} (x_j - c_j)}$$
Define $f(x_1, x_2, \cdots x_n) = \frac{\prod_{j} x_j-\prod_{j} c_j}{\prod_{j} (x_j - c_j)}$, noting that $f(1,1, \cdots, 1)$ is exactly the LHS above.
It thus suffices to show that for each index $i$, $\frac{\partial}{\partial x_i} f(x_1, x_2, \cdots x_n) \leq 0$ in the polytope $[0,1]^n \cap \bigcap_{i=1}^{n} \{x \in \mathbb{R}^n: x_i > c_i\}$.
By the quotient rule, $$\frac{\partial}{\partial x_i} f(x_1, x_2, \cdots x_n) \leq 0 \Longleftrightarrow$$$$ \left(\prod_{j \neq i} x_j\right)\prod_{j} (x_j - c_j) - \left(\prod_{j } x_j - \prod_{j} c_j\right) \prod_{j \neq i} (x_j - c_j) \leq 0 \Longleftrightarrow$$$$  \prod_{j \neq i} (x_j - c_j)\left((x_i - c_i)\prod_{j \neq i} x_j -  \left(\prod_{j } x_j - \prod_{j} c_j\right) \right) \leq 0 \Longleftrightarrow$$$$ (x_i - c_i)\prod_{j \neq i} x_j -  \left(\prod_{j } x_j - \prod_{j} c_j\right) = \prod_{j} c_j - c_i \prod_{j \neq i} x_j \leq 0$$
where the last equality is clear since $x_j > c_j$ for each $j$.
