show that $x^2e^x-\ln{x}-1>0$ 
For $x>0$ show that
  $$x^2e^x-\ln{x}>1$$

x^2e^x-lnx-1
if $x\ge 1$,we have
$$x^2e^x>2x^2>x\ge\ln{x}+1$$
But  for $0<x<1$ then It is hard.so How prove it?
 A: Rewrite the inequality to be proved as
$$\left(1+u\over2\right)^2e^{(1+u)/2}-\ln\left(1+u\over2\right)-1\gt0$$
for $-1\lt u\lt1$.  This can be rewritten as
$$(1+u)^2\sqrt ee^{u/2}-4\ln(1+u)+4\ln2-4\gt0$$
Now $e^{u/2}\ge1+{u\over2}$, so 
$$(1+u)^2\sqrt ee^{u/2}\ge(1+2u)\sqrt e\left(1+{1\over2}u\right)\ge\left(1+{5\over2}u\right)\sqrt e$$  
Also, $\ln(1+u)\le u$, so 
$$-4\ln(1+u)\ge-4u$$
Putting these together, we have, for $|u|\le1$,
$$\begin{align}
(1+u)^2\sqrt ee^{u/2}-4\ln(1+u)+4\ln2-4
&\ge\sqrt e+4\ln2-4+\left({5\over2}\sqrt e-4\right)u\\
&\ge\sqrt e+4\ln2-4-\left|{5\over2}\sqrt e-4\right|
\end{align}$$
It remains to compute $\sqrt e+4\ln2-4\approx0.4213$ while ${5\over2}\sqrt e-4\approx0.1218$.
There might be some slicker way to do this that avoids the messy numerics at the end.  I'd like to see it.
Added later:  Oh, here's a way to avoid the messy numerics.  Note first that $e\gt2.56=1.6^2=(8/5)^2$, so ${5\over2}\sqrt e-4$ is positive.  This simplifies things to showing $4\ln2\ge{3\over2}\sqrt e$.  If we now allow ourselves to know that $\ln2\gt{2\over3}$, it suffices to show that ${16\over9}\gt\sqrt e$, or ${256\over81}\gt e$, which is clear, since ${256\over81}\gt3\gt e$.
If you want a quick proof that $\ln2\gt{2\over3}$, note that that inequality is equivalent to $8\gt e^2$, which follows from $8\gt{196\over25}=\left(14\over5\right)^2=2.8^2\gt e^2$.
And just to leave no numerical stone unturned, here's why we can confidently say that ${14\over5}\gt e\gt{64\over25}$:
$$e\gt1+1+{1\over2}+{1\over6}=2+{2\over3}\gt2+{14\over25}={64\over25}$$
since $2\cdot25\gt3\cdot14$, and
$$e^{-1}\gt1-1+{1\over2}-{1\over6}+{1\over24}-{1\over120}={11\over30}\gt{5\over14}$$
since $11\cdot14\gt5\cdot30$.
A: There's an ad-hoc solution at least going something like this:
You can use taylor expansion to show that (or the convexity of the function). Consider the function $f(x) = x^2e^x-\ln x - 1$ and you have:
$$f'(x) = (x^2+2x)e^x - 1/x$$
$$f''(x) = (x^2+4x+4)e^x + 1/x^2 = (x+2)^2e^x + 1/x^2$$
We see immediately that $f''(x)\ge 0$. This means that the expansion around any point we will have:
$$f(a+h) = f(a) + f'(a)h + f''(\xi)h^2/2 \ge f(a) + f'(a)h$$
This means that it will be above every of it's tangent. Now one only needs to pick a tangent sufficiently close to the right of it's minimum such that either of them are positive which will prove that $f(x)$ is positive (since it's above both tangents).
The rest is ugly computations. You have that $f(1/2) > 1/10$ and that $0<f'(1/2)<1/10$ (you can find this by Taylor expansion - or if you're lazy by using calculator). This means that we have:
$$f(x) = f(1/2 + x-1/2) \ge f(1/2) + f'(1/2)(x-1/2)$$
Which means:
$$f(x)\begin{cases}
\ge 1/10 + 1/10 (x-1/2) = (1+2x)/10 \ge 0 & \text{when } x\le 1/2
\\ \ge 1/10 \ge 0 & \text{when } x \ge 1/2
\end{cases}$$
