The logarithmic inequality $\ln^q (1+x) \le \frac{q}{p} x^p \quad (x \ge 0, \; 0 < p \le q)$ $$\ln^q (1+x) \le \frac{q}{p} x^p \quad (x \ge 0, \; 0 < p \le q)$$
For $p=q$ this reduces to the familiar $\ln(1+x) \le x$. Otherwise I haven't had much success in proving it. General suggestions would be appreciated.

put $p=\frac{1}{\ln x}$ and when $x > e$ then $p <1$ and we arrive at 
$\ln^q(x+1) \leq  q \ln x  * x^{p}$ which is $\ln^q (x+1) \leq q \ln x *e$
when $q \geq 2$ we get that $\ln^q(x+1) \leq e q \ln x$
Because  $\ln^q(x+1)\geq \ln^q x \geq e q \ln x$ divide by $\ln x$
We get that $\ln^{q-1} x \geq e q $ and since $q \geq 2$
then we have that $\ln^{q-1} x \geq \ln x \geq e q$ which will be true whenever 
$\ln x \geq e q$ exponent -ate  both sides we get that 
$x \geq e^{e q}$ for example if we put $q=3$ then the inequality will be false for all $ x \geq e^{3e} \approx 3480.2$
