TITLE: Prove with use of derivative
QUESTION [5 upvotes]: How to prove this inequality using derivative ?
For each  $x>4$ , 
$$\displaystyle \sqrt[3]{x} - \sqrt[3]{4} < \sqrt[3]{x-4} $$

REPLY [6 votes]: Let $f(x)= x^{1/3} - 4^{1/3} - (x-4)^{1/3}$
Note that $f(x)=0$ for $x=4$ (which is just outside its domain)
Now, taking the derivative with respect to $x$,
$f'(x)=(1/3)x^{-2/3} -(1/3)(x-4)^{-2/3}$
The term on the right hand side is negative for $x>4$
Thus, the funtion $f(x)$ is a decreasing function for $x>4$.
So, $f(x)<0$ for all $x>4$, which yields the desired inequality.

REPLY [3 votes]: If $f(x)=x^\frac{1}{3}$ we must show that $$ f(x)  < f(x-4) + f(4)
$$
Note that for any $a,\ b>0$ if we have $f(a+b)< f(a) + f(b)$, then
we are done. Note that this is equivalent to $$ f(a+b)-f(a) < f(b) -
f(0) $$
In further this is
equivalent to $$ \int_a^{a+b} f'(x)dx < \int_0^b f'(x) dx $$
So $f'(x) = \frac{1}{3} \frac{1}{x^\frac{2}{3}} $ is decreasing so
that in the above right integral is large