Here is a counterexample of size 38.

Let $m=8$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i<j\le m\}$$ where $0<a_i<b_{jk}<1$ whenever $i$ is distinct from $j$ and $k$.

The cardinality of $P$ is $n=1+m+\binom{m}{2}+1=38$.

The join-irreducible elements are only the $a_i$ and $0$, since $b_{ij}=\bigvee\{a_k:k\ne i, k\ne j\}$.

The $a_i$ each have $\binom{m}{2}-m=20>19=n/2$ elements above them.

Here is a counterexample of size 23.

Let $m=6$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i<j\le m\}$$ where $0<a_i<b_{jk}<1$ whenever $i$ is distinct from $j$ and $k$.

The cardinality of $P$ is $|P|=m+2+\binom{m}{2}=6+2+15=23$.

The join-irreducible elements are only the $a_i$ and $0$, since each $b_{ij}=\bigvee\{a_k:k\ne i, k\ne j\}$.

Each $|[a_i,1]|=2+\binom{m}{2}-(m-1)>|P|/2$ as long as $$2+\binom{m}2 - (m-1) > \frac12\left(m+2+\binom{m}2\right)$$ $$4+\binom{m}2>3m$$ which is true for $m=6$ but not for $m=5$.