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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?

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  • $\begingroup$ This seems to be much more like Bernoulli's inequality $\endgroup$ – Rohan Shinde Feb 6 '18 at 2:35
  • $\begingroup$ It is Bernoulli's inequality @Manthanein; the strict version. $\endgroup$ – Angel Politis Feb 6 '18 at 2:37
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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 $$ \alpha > 1 \Longrightarrow (1+k)^{\alpha-1} > 1 \Longrightarrow (1+x)^\alpha > 1 + \alpha x \, , \\ 0 < \alpha < 1 \Longrightarrow (1+k)^{\alpha-1} < 1 \Longrightarrow (1+x)^\alpha < 1 + \alpha x \, . $$

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  • $\begingroup$ Thanks a lot for the answer @MartinR. I knew my solution was on the right track, but something was amiss 😊 $\endgroup$ – Angel Politis Feb 6 '18 at 22:41

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