Proof for inequality $1 - \frac{\ln^2n}{n} \ge \left(1 - \frac{\ln n}{n} \right)^{k \ln n}$ I would like to prove that
$$ f(n) \ge g_k(n) $$
with
$$f(n) = 1 - \frac{\ln^2n}{n}$$ and $$g_k(n) = \left(1 - \frac{\ln n}{n} \right)^{k \ln n}$$
for $n, k \in \mathbb{R}$ with $n > 1$ and $k > \gamma$ where $\gamma$ seems to be between $1$ and $2$.
It is obvious that at $n = 1$ both sides are equal and numerically the inequality seems to hold for large enough $k$ but I have not been able to come up with a proof.
Numeric Visualisation

Here are some plots of the difference between the left and the right side of the inequality:
$$d_k(n) = f(n) - g_k(n) = 1 - \frac{\ln^2n}{n} - \left(1 - \frac{\ln n}{n} \right)^{k \ln n}$$
for $k = 2$ the inequality seems to hold:

for $k = 1$ it does not hold:

for $k = 1.2$ it is a bit strange:

 A: $$1 - \frac{\ln^2n}{n} \ge \left(1 - \frac{\ln n}{n} \right)^{k \ln n}$$
$$\left( 1 - \frac{\ln^2n}{n}\right)^{\frac{n}{\ln^2n}} \ge \left(1 - \frac{\ln n}{n} \right)^{\frac{n}{\ln n}k}$$
let
$$f\left(\frac{n}{\ln^2n}\right)\geq f\left(\frac{n}{\ln n}\right)^k$$
with "f" monotonically increasing
$$f(x)=(1-\frac1x)^x \leq \frac1e$$
thus
$$\ln \left[ f\left(\frac{n}{\ln^2n}\right)\right]\geq k\ln \left[ f\left(\frac{n}{\ln n}\right)\right]$$
that is

$$k \geq MAX\left(\frac{\ln \left[ f\left(\frac{n}{\ln^2n}\right)\right]}{\ln \left[ f\left(\frac{n}{\ln n}\right)\right]}\right)= M\approx 1.25447$$


http://www.wolframalpha.com/input/?i=plot+(ln((1-log%5E2x%2Fx)%5E(x%2Fln%5E2x))%2Fln((1-lnx%2Fx)%5E(x%2Flnx)))+x%3D1..250
http://www.wolframalpha.com/input/?i=local+max+(ln((1-log%5E2x%2Fx)%5E(x%2Fln%5E2x))%2Fln((1-lnx%2Fx)%5E(x%2Flnx)))
NOTE
the existence of MAX is guarantee by Weierstrass Theorem since:


*

*for $n=e$ the ratio is $1$

*for $n=2e$ the ratio is $>1 $

*for $n\to +\infty$ the ratio $\to 1$

