I am currently working on an olympiad inequality problem, and I am feeling stuck. The problem is this:
Prove the following inequality $$\frac{1}{\sqrt[2017]{2018}}+\frac{1}{2017}(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2018})< 1$$
This is my attempt to solve:
We have $$2018^{1/2017} > 2018^{1/2018} > 2048^{1/2048} = 2^{11/2048}$$
Using $\frac{1}{k} < \ln k - \ln(k - 1)$ for all $k \ge 2$, we have $$\frac{1}{3} + \cdots + \frac{1}{2018} < \ln(2018) - \ln 2 < \ln 2048 - \ln 2 = 10\ln 2.$$
It suffices to prove that $$2^{-11/2048}+\frac{1}{2017}(1/2 + 10\ln 2) < 1$$ or $$\ln\left(1 - \frac{1}{2017}(1/2 + 10\ln 2)\right) + \frac{11}{2048}\ln 2 > 0.$$
I assumed the inequality $\frac{1}{k} < ln k - ln(k-1)$, and I verified it worked on desmos. But I am not familiar with logs and I don't know how to prove it.
Now at the last step is where I am stuck. I don't know how to continue and solve this inequality. Can someone give me some help and tell me how to solve it?
Thanks for help...
Not really related to your doubt but a slightly simple way of going about it:
$\frac{1}{2018^{1/2017}} + \frac{1}{2017} (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{2018}) < 1 \\ \frac{1}{2018^{1/2017}} + \frac{1}{2017} (\frac{2-1}{2} + \frac{3-2}{3} + \frac{4-3}{4} + ... + \frac{2018-2017}{2018}) < 1 \\ \frac{1}{2018^{1/2017}} + 1 - \frac{1}{2017} (\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + ... + \frac{2017}{2018}) < 1 \\ \frac{1}{2018^{1/2017}} <\frac{1}{2017} (\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + ... + \frac{2017}{2018}) \\$
And this is true using AM-GM inequality