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In the book of The Elements of Analysis by Bartle, at page 214, it is given that

To show that $(1+x)^n \geq 1 + nx$ for $n \geq 1$ and $n\in \mathbb{R}$, let $$f(x) = (1+x)^r$$ so that $$f'(x) = r (1+x)^{r-1}.$$ If $-1 < x < 0$, then $f'(x) < r$, while if $x>0$, then $f'(x) > r$.If we apply MVT to both of those cases we obtain $$(1+x)^r \geq 1 + rx,$$ when $1+x> 0$ and $r \geq 1$.

However, I couldn't understand what does the author means by saying "applying the MVT to both of those cases". I do not understand how do we get the conclusion from those observations. I would appreciate if someone can explain to me the procedure.

Note that I have seen this question, but I'm particularly interested in this particular solution.

Our
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3 Answers3

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So we have $r\geq 1$. I do the case $x>0$ and you do the other case. We apply the MVT to the interval $[0,x]$. You can verify the conditions are satisfied. Hence by the MVT there is $c\in (0,x)$ such that \begin{align} \frac{(1+x)^r-1}{x}= f'(c) \end{align} But we have $f'(c) = r(1+c)^{r-1}\geq r$ so: \begin{align} \frac{(1+x)^r-1}{x}= f'(c) \geq r \end{align} Multiply both sides with $x$ (note $x>0$): \begin{align} (1+x)^r-1 \geq rx \end{align} Add $1$ on both sides \begin{align} (1+x)^r \geq rx+1 \end{align} The other case is for you.

Shashi
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The mean value theorem says:

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that $$\frac{f(b)-f(a)}{b-a} = f'(c)$$

Put another way, the conclusion is that $$ f(b) = f(a) + f'(c)(b-a) $$ for some $c$ between $a$ and $b$.

Assume $x>0$, and apply MVT to $f(x) = (1+x)^n$, $a=0$, and $b=x$. You don't know what $c$ is, other than that $0 < c < x$, but at least you know $f'(c) > n$. So $$ f(x) = f(0) + f'(c) x > 1 + nx $$ The case $x<0$ is treated similarly.

Editorial note: The Mean Value Theorem is The Most Important Theorem in Calculus!

Matthew Leingang
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Note that $f(x) = (1+x)^n$ is a convex function on $[-1,\infty)$.

Since $f'(x) = n(1+x)^{n-1}$, the tangent line on $x=0$ is $$ y = nx+1 $$ As $f$ is convex on $[-1, \infty)$, it satisfy [ref] that $$ f(0) + f'(0)(x - 0) \le f(x) $$ in other words $$ nx + 1 \le (1+x)^n \quad \text{where} \quad [-1, \infty).$$

induction601
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