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

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
