Show that $1<\sqrt{1+x^3}<1+x^3$ for $x>0$ The problem:

Show that $1<\sqrt{1+x^3}<1+x^3$ for $x>0$

How do I solve this using the Fundamental Theorem of Calculus
 A: You don't need the  Fundamental Theorem of Calculus to prove this proposition. Since $x>0$, we get $1<1+x^3$ and $\sqrt{1}<\sqrt{1+x^3}$. We know that $1+x^3>1$ and $t<t^2$ if $t>1$ so $\sqrt{1+x^3}<1+x^3$.
A: As other people have pointed out, you certainly don't need the Fundamental Theorem of Calculus to prove these inequalities, but if you do want to use it, here's a way to get one of them at least:
If $F(t)=\sqrt{1+t^3}$, then $F'(t)=3t^2/2\sqrt{1+t^3}$, which is clearly postive for any $t\not=0$ in its domain, which means that it has a positive integral over any interval of positive length.  Thus, by FTC,
$$0\lt\int_0^x F'(t)dt = F(x)-F(0)=\sqrt{1+x^3}-1$$
for $x\gt0$, which gives the first inequality.  
You can get the other inequality by using $F(t)=t^3-\sqrt{1+t^3}$, but showing $F'(t)\gt0$ requires showing $1\lt2\sqrt{1+t^3}$ for $t\gt0$, which I suppose you can justify because you've already shown that $1\lt\sqrt{1+x^3}$ for $x\gt0$.
A: consider $f(x) = \sqrt(1+x^3) - 1$ and $g(x) = 1 + x^3 - \sqrt(1+ x^3)$ prove that both $f$ and $g$ are positive if $x$ is positive. 
to prove this, see that $f(0) = g(0) = 0$. now check that both $f$ and $g$ are strictly increasing by looking at its derivative.
This method you can apply in case of most of such inequality.
A: you just want to show that $1<\sqrt{x}<x$ for all $x>1$, that is, $x<x^2$ for all $x>1$. Note that $x^2-x=x(x-1)>0$, you can prove the inequality.
