Proving that $f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$ is strictly positive for odd integers $n\geq 5$ 
Let $n\geq 5$ be an odd integer. I'd like to prove that the function $$f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$$ is strictly positive on $\mathbb{R}$.

(I already figured out that the numerator is divisible by the denominator, so $f_n(x)$ is in fact a polynomial of degree $n-3$.
I tried to find the minimum by differentiation, but I have difficulty finding the location and the value of the minimum. Another thing I know is that $f_n(x)$ is palindromic, but it doesn't seem to give any inspirations. Does anyone have ideas?
Thanks in advance!

First, we have
$$f_5(x) = 5(x^2 + x + 1) > 0, \, \forall x \in \mathbb{R}.$$
Second, for $k\ge 2$, let
\begin{align*}
 g_k(x) &:= f_{2k+3}(x) - f_{2k+1}(x)\\[5pt]
 &= \frac{(x + 1)^{2k + 3} - (x + 1)^{2k + 1} - x^{2k + 3} + x^{2k + 1}}{x(x + 1)}\\
 &= \frac{(x + 1)^{2k}[(x + 1)^3 - (x + 1)] - x^{2k}(x^3 - x)}{x(x + 1)}\\
 &= \frac{(x + 1)^{2k}x(x + 1)(x + 2) - x^{2k}x(x - 1)(x + 1)}{x(x + 1)}\\
 &= (x + 1)^{2k}(x + 2) - x^{2k}(x - 1).
\end{align*}
We can prove that $g_k(x) \ge 0$ for all $x\in \mathbb{R}$.
If $-2 < x < 1$, clearly $g_k(x) \ge 0$.
If $x \le -2$, we have $g_k(x) = (1 - x)(-x)^{2k} - (-x - 2)(-x - 1)^{2k} \ge 0$ since $1 - x > - x - 2 \ge 0$
and $- x > - x - 1 \ge 1$.
If $x \ge 1$, we have $g_k(x) \ge 0$
since $x + 1 > x \ge 1$
and $x + 2 > x - 1 \ge 0$.
We are done.
