Let real $x \geq 0 $ and real $p > 2$.
Let $ f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1$.
Show that, for the given range of $x$ and $p$, $f(x)=0$ only for $x=0$ and $x=1$.
Since $f(x)$ is not convex, I find it difficult to show that $f(x) < 0$ for $0<x<1$ and $f(x) > 0$ for $x>1$.
EDIT (generalization): the same zeroes hold for all $p>0$ ; $p \neq 1$ , $p \neq 2$.
We will show that $f(x) > 0$ for $x>1$ and $f(x) < 0$ for $0<x<1$.
Let's start with the first one, $f(x) > 0$ for $x>1$.
For real $p > 2$, we can use Bernoulli's inequality (which holds for real exponents!):
$(1 + w)^r \ge 1 + r w $, for real $w > -1$ and real $r > 1$.
This gives for the first exponent in $f(x)$:
$(x+1)^{p-1} = x^{p-1} (1+1/x)^{p-1} \geq x^{p-1} (1 + \frac{p-1}{x})$
Inserting into $f(x)$ gives
$$f(x) = (x - 1)(x + 1)^{p - 1} - x^p + 1 \\\ge (x - 1)x^{p-1} (1 + \frac{p-1}{x}) - x^p + 1 \\= x^{p-1} (p-2) - x^{p-2} (p-1) + 1 = h(x)$$
with equality for $x=1$. Now note that $h(x=1) = 0$. Further,
$h'(x) = (p-2)(p-1)x^{p-3} (x-1)$
so for $x>1$, $h'(x) > 0$ everywhere. Hence $h(x) \geq 0$ with equality for $x=1$. This establishes that for $x\geq 1$, $f(x) \geq 0$ with equality for $x=1$.
For the second case $0<x<1$, we will see now that this can be reduced to the first case $x>1$, so indeed there is nothing to do.
Let $y =\sqrt x$. Then the function becomes:
$$ f(y = \sqrt x) = y^p \left[ (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} \right] $$
We have shown above that for $y>1$, $f(y) >0$. I.e. we have that the following inequality holds:
$$ g(y) = (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} > 0 \qquad \bf\text{[ineq. 1]} $$
Now, we want to show that for $0<y<1$, $f(y) <0$. Here, we need to show
$$
g(y) =(-y + \frac1y)(y + \frac1y)^{p-1} -[- y^p + \dfrac{1}{y^p} ]> 0
$$
Now replacing $y$ by $1/y$ we get $y > 1$ and, in the replaced variable,
$$
\tilde g(y) = (y - \frac1y)(y + \frac1y)^{p - 1} - y^p + \dfrac{1}{y^p} > 0
$$
So this is the same condition as before in ineq. 1, which has already been proven.
$\qquad \qquad \Box$
The term $x^p-1$ for $p \geq 1$ is zero for $x=1$ and therefore it has the factor $x-1$. It can easily be checked that this results in: $$ x^p - 1 = (x-1) \sum_{k=0}^{p-1} x^k $$ we can therefore rewrite the function $f(x)$ as $$ f(x) = (x-1) \left[ (x+1)^{p-1} - \sum_{k=0}^{p-1} x^k\right] $$ where we can now use the binomial expansion on the term $(x+1)^{p-1}$ to obtain $$ f(x) = (x-1) \left[ \sum_{k=0}^{p-1} \binom{p-1}{k} x^k - \sum_{k=0}^{p-1} x^k\right] $$ and if we combine the two sums we find $$ f(x) = (x-1) \sum_{k=0}^{p-1} \left[ \binom{p-1}{k} - 1 \right] x^k $$ The sum is a polynomial (of degree $p-2$ because the terms proportional to $x^0$ and $x^{p-1}$ vanish) with positive coefficients and hence has a positive value for all $x>0$. Therefore $f(x)$ has only a single root on $x>0$, which is $x=1$ and together with $f(0)=0$ concludes the proof.
