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!
 A: 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.
A: I will treat $f$ as  polynomial of degree $n-3$.
If $f$ takes both positive and negative values then it must vanish at some point. But $(x+1)^{n}=x^{n}+1$ has no  solution in the real line other than $-1$ and $0$: Its derivative is strictly positive in $(-\frac 1 2 , \infty)$ and strictly negative in $(-\infty, -\frac 1  2 )$. So it attains its minimum at $-\frac  1 2$ and there are exactly two values at which it vanishes. Of course, these are $0$ and $-1$. Now $f_n(0)\neq 0$ since $x^{2}$ is not  factor of the numerator and $f_n(-1)\neq 0$ since $(1+x)^{2}$ is not  factor of the numerator. Hence, $f$ does not change sign. Also, $f(1)>0$ so $f(x) >0$ for all $x$.
A: We need to prove that for all $x$ the inequality
$$
\frac{(x+1)^n-x^n-1}{x(x+1)}>0
$$
holds. Consider three cases:

*

*$x>0$. Then, $x(x+1)>0$ and
$$
(x+1)^n-x^n-1=\sum_{k=1}^{n-1}\binom{n}{k}x^k>0.
$$

*$x<-1$. Denote $y=-x-1$, then $y>0$ and $x(x+1)>0$ and we need to check that
$$
\frac{(x+1)^n-x^n-1}{x(x+1)}=\frac{-y^n+(y+1)^n-1}{y(y+1)}>0.
$$
This is similar to the first case.

*$-1<x<0$. Denote $t=-x\in(0,1)$, Then, we have
$$
(x+1)^n-x^n-1=(1-t)^n+t^n-1<(t+(1-t))^n-1=0.
$$
Thus, $(x+1)^n-x^n-1<0$ while $x(x+1)<0$.

