Proving two inequalities in calculus I'm trying to prove this two statements:
$$1) \quad \forall\,x\geq 1, \quad \left|\arctan(x)-\frac{\pi}{4}-\frac{x-1}{2}\right|\leq \frac{(x-1)^2}{2}; \\
2)\quad\textrm{If}\;e <a<e^{2}\implies a^{\ln a}-e<4e^3(a-e)\qquad\quad ;
$$
For the second inequality, before trying to solve I saw some things in the statement.
If $a>e\implies  \ln(a)>\ln(e)=1\implies a^{\ln a}-e=e^{(\ln a)^{2}}-e>e^{(\ln(e))^{2}}-e=0\implies a^{\ln a}-e>0.$
And... this is all I got. I don't know what to do.
For the first inequality, I've been searching for some similar questions. The answers always comes from a clever function definition and applying the mean value theorem. Is that the way to solve this problem?
I know that I didn't get so close, but any help would be appreciated. 
 A: Here's a proof of the first.
Since
$\arctan(x)-\arctan(y)
=\arctan(\dfrac{x-y}{1+xy})
$
and
$\dfrac{\pi}{4}
=\arctan(1)
$,
$\arctan(x)-\frac{\pi}{4}
=\arctan(x)-\arctan(1)
=\arctan(\dfrac{x-1}{1+x})
$
the first inequality is
$\left|\arctan(\dfrac{x-1}{1+x})-\dfrac{x-1}{2}\right|\leq \dfrac{(x-1)^2}{2}
$.
Let
$y
= \dfrac{x-1}{1+x}
=1-\dfrac{2}{1+x}$
so
$0 \le y \le 1$.
$y+yx = x-1,\\
y+1 = x(1-y),\\
x = \dfrac{1+y}{1-y},\\
x-1 
= \dfrac{1+y}{1-y}-1
=\dfrac{1+y-(1-y)}{1-y}
=\dfrac{2y}{1-y},
$
or
$\dfrac{x-1}{2} 
=\dfrac{y}{1-y}
$.
This becomes
$\left|\arctan(y)-\dfrac{y}{1-y}\right|
\leq \dfrac{2y^2}{(1-y)^2}
$.
We have 
$y
\ge \arctan(y)
\ge y-\dfrac{y^3}{3}
$
(to prove this, integrate
$1 \ge \dfrac1{1+t^2}
\ge 1-t^2$
from $0$ to $y$)
so
$\begin{array}\\
f(y)
&=\dfrac{y}{1-y}-\arctan(y)\\
&\ge \dfrac{y}{1-y}-y\\
&=\dfrac{y-y(1-y)}{1-y} \\
&=\dfrac{y^2}{1-y}\\
&\ge 0\\
\text{and}\\
g(y)
&=\dfrac{y}{1-y}- \dfrac{2y^2}{(1-y)^2}-\arctan(y)\\
&\le \dfrac{y}{1-y}- \dfrac{2y^2}{(1-y)^2}-(y-\dfrac{y^3}{3})\\
&= \dfrac{(y - 3) y^2 (y^2 + y + 1)}{3 (y - 1)^2}
\qquad\text{(according to Wolfy)}\\
&\le 0
\qquad\text{since } y < 3\\
\end{array}
$
Therefore
$0
\le \arctan(y)-\dfrac{y}{1-y}
\le \dfrac{2y^2}{(1-y)^2}
$.
A: For the second inequality, let $a=eb$ with $1\lt b\lt e$. Note that
$$a^{\ln a}=(eb)^{\ln(eb)}=(eb)^{1+\ln b}=e^{1+\ln b}b^{1+\ln b}=ebb^{1+\ln b}=eb^{2+\ln b}$$
so the inequality to prove becomes
$$b^{2+\ln b}-1\lt4e^3(b-1)$$
for $1\lt b\lt e$.  Now $1\lt b\lt e$ implies $b^{\ln b}\lt b^{\ln e}=b$, so
$$b^{2+\ln b}-1\lt b^3-1=(b-1)(b^2+b+1)\lt(b-1)(e^2+e+1)\lt4e^3(b-1)$$
where the final inequality, $e^2+e+1\lt4e^3$, is obvious.
