Prove that $b^3 \log(b) - a^3\log(a) \le 4e^2(b-a)$ I have no idea where to start here and I am not even sure what topic this actually is a part of. Any help would be great.

Prove that for all $a,b \in [1,e]$ we have: $$b^3 \log(b) - a^3\log(a) \le 4e^2(b-a)$$ 

Thanks in advance
 A: Hint
Take the function
$$f(x)=x^3\log x$$
then, $$f'(x)=3x^2\log x+x^2=x^2(3\log x+1)$$
by, Mean Value Theorem, there is $c\in (b,a)$ such that
$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$
Take $c\in [1,e]$ and show that $f'(c)\le 4e^2$.
A: Let $g(x)= x^2(3\log x+1)$, then $g'(x) = 2x\log x +5x$, so $g'(x)>0$ for all $x\in [1,e]$ and so $g$ is increasing function, so $g(x)\leq g(e) = 4e^2$, for all $x\in [1,e]$. 
Now since $x^2(3\log x+1)\leq 4e^2$ for all $x\in [1,e]$ then we have $$\int _{a}^bx^2(3\log x+1)dx\leq \int _{a}^b 4e^2dx $$
so 
$$ b^3 \log(b) - a^3\log(a) \le 4e^2(b-a)$$
A: Another way to "prove" it is to realize that $y = f(x)=x^3\ln x - 4e^2x$ on $[1,e]$ has $y' = x^2+3x^2\ln x- 4e^2\le 0, y''= 5x+6x\ln x>0\implies f(b) \le f(a), b \ge a$, and this is the conclusion.
A: The desired inequality is equivalent to
$$4e^2a-a^3\log a\le4e^2b-b^3\log b$$
if $1\le a\le b\le e$. (Note, in particular, the OP seems to be making the tacit assumption $a\le b$.) This, in turn, is equivalent to asserting that
$$f(x)=4e^{2+x}-e^{3x}x$$
is increasing for $0\le x\le 1$.  But it's easy to see that
$$f'(x)=4e^{2+x}-3e^{3x}x-e^{3x}=e^x(4e^2-(3x+1)e^{2x})\ge e^x(4e^2-4e^2)=0$$
with strict inequality holding for $x\lt1$. So $f$ is increasing, hence the desired inequality holds.
