# 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?

• This seems to be much more like Bernoulli's inequality – Rohan Shinde Feb 6 '18 at 2:35
• It is Bernoulli's inequality @Manthanein; the strict version. – Angel Politis Feb 6 '18 at 2:37

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 $$\alpha > 1 \Longrightarrow (1+k)^{\alpha-1} > 1 \Longrightarrow (1+x)^\alpha > 1 + \alpha x \, , \\ 0 < \alpha < 1 \Longrightarrow (1+k)^{\alpha-1} < 1 \Longrightarrow (1+x)^\alpha < 1 + \alpha x \, .$$