Without calculus, show that the sign of $(1+a)^{1+x}-a^{1+x}-1$ matches the sign of $x$, for any positive $a$ Let $a$ be a positive real constant. Consider the function
$$
f(x) = (1+a)^{1+x}-a^{1+x}-1
$$
We have for any choice of $a$, that $f(x) >0$ for $x>0$ and $f(x) <0$ for $x<0$, which can be proved using calculus. 
My question: 

How would a proof go without using calculus?


For positive $x$ we have
$$
\begin{align}
 \frac{1+a^{1+x}}{(1+a)^{1+x}} &= \left( \frac{1}{1+a}\right)^{1+x} + \left( \frac{a}{1+a}\right)^{1+x}\\
 &= 
\frac{1}{1+a} \left( \frac{1}{1+a}\right)^{x} +  \frac{a}{1+a}\left( \frac{a}{1+a}\right)^{x} \\
&< \frac{1}{1+a} + \frac{a}{1+a} = 1 \, .
\end{align}
$$
For negative $x$ the same holds with $<$ replaced by $>$.
