Question: Let $n\geq 1$. A contest has two rounds. You are automatically in the second round, while $n$ other people are in the first round. Each of the $n$ other people proceeds to the second round with probability $1/n$, independently. If there are $k$ people in the second round, you win the contest with probability $1/k$. Is it true that you win the contest with probability at least $1/2$, regardless of $n$?

For $n=1$ it's clear that the probability is exactly $1/2$.

For $n=2$, the probability that both other people join is $1/4$, and in this case you win with probability $1/3$. If one other person joins (probability $1/2$), you win with probability $1/2$. If no one else joins, you win with probability $1$. The probability that you win is then $$\frac14\cdot\frac13+\frac12\cdot\frac12+\frac14\cdot 1=\frac{7}{12}>\frac12.$$

For $n=3$, the probability that you win is 
$$\frac{1}{27}\cdot\frac{1}{4}+\frac{6}{27}\cdot\frac13+\frac{12}{27}\cdot\frac12+\frac{8}{27}=\frac{65}{108}>\frac12.$$

Answer: 
To answer your question, let $X\sim Binom(n,\frac 1 n)$. Then $$P(Win(n))=E\big(\frac 1 {1+X}\big)\geq\frac 1 {1+E(X)}=\frac 1 2$$ by Jensen's inequality (with equality only if $n=1$).

We can also integrate probability generating function for $X$ from $0$ to $1$ to yield $$P(Win(n))=\sum\_{k=0}^n\frac 1 {k+1}P(X=k)=\sum\_{k=0}^n\int\_0^1P(X=k)z^kdz=\int\_0^1 (1-\frac 1 n+\frac z n)^ndz=\frac n {n+1}-\frac{n-1}{n+1}\big(1-\frac 1 n)^n\to 1-e^{-1}$$
Let $p\_i$ be the probability that $i$ other people reach the second round. The expected number of other people who reach the second round is $1$. Hence $p\_0\ge1-p\_0-p\_1$, so your winning probability is at least

$$
p\_0+\frac12p\_1\ge\frac12p\_0+\frac12p\_1+\frac12(1-p\_0-p\_1)=\frac12
$$

(and this would hold even if you lost if two or more other people reach the second round).