Prove inequality using the Mean Value Theorem I'm trying to hone my problem-solving skills using the Mean Value Theorem and in one exercise, where $x \in (0, +\infty)$, I have to prove that:


*

*$(1+x)^a>ax+1$,   if $a > 1$.

*$(1+x)^a<ax+1$,   if $a \in (0, 1)$.


What I've tried:
I've tried to solve this problem using the function $f(t)=(1+t)^a$ in the closed set $[0,x]$ as follows:


*

*First, I calculated the derivative of $f$, which is $f'(t)=a(1+t)^{a-1}$.

*Then, I used the Mean Value Theorem:
$$
f'(k)={{f(x)-f(0)}\over{x-0}}={{(1+x)^a-1}\over{x}}
\Leftrightarrow
a(1+k)^{a-1}={{(1+x)^a-1}\over{x}}\\\Leftrightarrow
(1+x)^a=ax(1+k)^{a-1}+1
$$


The equation I found seems to be on the right track, so I decided based on instinct to examine the following cases:


*

*$a=1 \Rightarrow (1+x)^a=ax+1$

*$a>1 \Rightarrow (1+x)^a>ax+1$

*$a \in (0, 1) \Rightarrow (1+x)^a<ax+1$


Question:
My solution, and more specifically the part where my instinct kicks in, feels rather incomplete and rushed. Is there a better way to solve this problem using the Mean Value Theorem?

Your “instinct” is correct, and it requires only small additions to
make it a full proof.
The mean value theorem implies that for $x > 0$
$$
 (1+x)^\alpha = 1 + \alpha x (1+k)^{\alpha-1}
$$
for some $k \in (0, x)$. It is relevant that $k$ is strictly positive,
so that one can continue to argue
$$
 \alpha > 1 \Longrightarrow (1+k)^{\alpha-1} > 1 
 \Longrightarrow (1+x)^\alpha > 1 + \alpha x \, , \\
 0 < \alpha < 1 \Longrightarrow (1+k)^{\alpha-1} < 1 
 \Longrightarrow (1+x)^\alpha < 1 + \alpha x \, . 
$$
