Proof with Lagrange Remainder Theorem

Question

I'm trying to prove that $$1+ \frac{x}{3} - \frac{x^2}{9}<(1+x)^{1/3}<1+\frac{x}{3}$$ if $x>0$.

Using the Lagrange remainder theorem, I have that $$(1+x)^{1/3}= 1+ \frac{x}{3} - \frac{x^2}{9}+\frac{f^{(3)}(c)}{3!}x^3 $$ for some $c$ such that $0<c<x$. Then because $f^3(c)=\frac{10}{27}(1+c)^{-8/3}$, we have that $$\frac{f^{(3)}(c)}{3!}x^3 >0$$and thus the left half of my inequality follows. I'm having trouble with the right side.

Answer 1

You were on right track and the right inequality is much easier. We have $$f(x) = f(0) + xf'(0) + (x^{2}/2)f''(c)$$ for some $c \in (0, x)$. Putting $f(x) = (1 + x)^{1/3}$ we see that $f''(c) = (-2/9)(1 + c)^{-5/3} < 0$ for all $c > 0$. and hence $$f(x) < f(0) + xf'(0)$$ i.e $$(1 + x)^{1/3} < 1 + \frac{x}{3}$$

Answer 2

Let $g(x)=1+\frac{x}{3}-(1+x)^{1/3}$, then $g'(x)=\frac{1}{3}\left[1-\frac{1}{\left(1+x\right)^{2/3}}\right]$, so $g(x)>0$ for $x>0$ and $g$ is increasing for $x>0$, since $g(0)=0$ it follows $g(x)>0$ what means $$(1+x)^{1/3}<1+\frac{x}{3}\;\;\text{for}\;\;x>0.$$