Is the inequality $x^4 (\ln (x^4) - 1) \ge x - e$, with $x>0$, true? The inequality
$$
x^4 (\ln (x^4) - 1) \ge x - e \qquad (x>0)
$$
is giving me some trouble. It is fairly crude, with the straight line in the right member lying well below the graph of the left member. Straightforward differentiation seems unpromising unless I've missed something. Potential proof approaches would be much appreciated. 
 A: It is equivalent to $x(\log x-1)\geq x^{1/4}-e$, and this is a fairly simple inequality to prove, since the LHS is a convex function while the RHS is a concave function on $\mathbb{R}^+$. For instance,
$$ g(x)=x^{1/4}-e \leq g'(1)(x-1)+g(1) = \tfrac{1}{4}(x-1)+(1-e) $$
while by setting $f(x)=x(\log x-1)$ we have $f'(x)=\log x$, from which
$$ f(x) \geq f'(e^{1/4})(x-e^{1/4})+f(e^{1/4}) = \tfrac{1}{4}(x-e^{1/4})-\tfrac{3}{4}e^{1/4} $$
and the claim follows from
$$ e \geq \tfrac{3}{4}+e^{1/4} $$
which is quite crude and rather simple to prove.
A: Let $f(x) = x^4(\ln(x^4)-1)$. Then
$$f'(x) = 4x^3(\ln(x^4)-1)+x^4\left(\frac{4}{x}\right) = 4x^3\ln(x^4) $$
from which we see thar $f'(x)<0$ for $0<x<1$. Since $f(1) = -1>1-e$, it follows that $f(x)>x-e$ for all $0<x<1$ since $f$ is decreasing on $(0,1)$, while $x-e$ is increasing.
In addition, since $f'(x)>0$ for $x>1$, it follows that for $x\in[1,e^{1/4}]$ we have
$$ f(x)\ge f(1) = -1 > e^{1/4}-e \ge x-e,$$
thus showing that $f(x)>x-e$ on $[1,e^{1/4}]$. Finally, for $x>e^{1/4}$ we have
$$ f'(x) = 4x^3\ln(x^4)> 4e^{3/4}\ln((e^{1/4})^4) > 1, $$
and since $f(e^{1/4})> e^{1/4}-e$, it follows that $f(x)>x-e$ on $(e^{1/4},\infty)$ as well.
A: For $x>0$ we'll prove a stronger inequality:
$$x^4 (\ln x^4 - 1) \ge x - \frac{6154}{2401}$$ or
$f(x)>0$, where
$$f(x)=\ln{x}-\frac{x^4+x-\frac{6154}{2401}}{4x^4}.$$
Indeed,
$$f'(x)=\frac{(7x-8)(1372x^3+1568x^2+1792x+3077)}{9604x^5},$$
which gives $x_{min}=\frac{8}{7}$ and since
$$f\left(\frac{8}{7}\right)=\ln\frac{8}{7}-\frac{343}{8192}>0,$$
we are done!
