Solution
= Solution
Use natural logarithms. A $K_t$ <graph minor> is represented by $t$ disjoint nonempty <branch sets of a graph minor>, each inducing a <connected graph>, with an <edge> between every pair. The <complete graph minor density threshold> is defined over nonempty finite <graphs> by
$$
c(t)=\inf\{c:\text{for every nonempty finite }G,\ e(G)\geq c|G|\Longrightarrow G\succ K_t\}.
$$
It uses $e(G)/|G|$, half the average <degree of a vertex>; the nonempty convention excludes the vacuous density inequality for the empty <graph>. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.
\b[Probabilistic lower bound.] Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the <binomial random graph> $G(N,1/2)$. Its <edge> count has mean $N(N-1)/4$. The <Chernoff bound> implies
$$
\mathbb P(e(G)\geq N^2/8)\longrightarrow1.
$$
We show that the <probability> of a $K_t$ <graph minor> tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ <vertices> one of $t$ branch labels or an unused label. There are at most $(t+1)^N$ assignments.
For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\leq\sqrt{\log t}/2$. Between any two such small sets $A,B$, the <probability> of no <edge> is
$$
2^{-|A||B|}\geq \exp\left(-\frac{\log2}{4}\log t\right)\geq t^{-1/4}.
$$
The absence or presence of <edges> between different pairs of branch sets depends on disjoint sets of random <edges>, so these events are independent. For this assignment the <probability> that all small pairs are adjacent is at most
$$
(1-t^{-1/4})^{\binom{\lfloor t/2\rfloor}{2}}
\leq\exp\left(-\binom{\lfloor t/2\rfloor}{2}t^{-1/4}\right).
$$
The <union bound> over assignments is therefore at most
$$
\exp\left(N\log(t+1)-\binom{\lfloor t/2\rfloor}{2}t^{-1/4}\right)\longrightarrow0,
$$
since the positive term is $O(t(\log t)^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a <graph> with no $K_t$ <graph minor> and with $e(G)/N\geq N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\geq t\sqrt{\log t}/64$ for large $t$,
$$
\boxed{c(t)\geq\beta t\sqrt{\log t}\quad\text{with }\beta=1/64.}
$$
\b[The auxiliary density lemma contains a typo.] Both the original PDF and the TeX print $e(G)\leq11k|G|$ in the hint. This cannot imply its conclusion: for $k\geq1$, an edgeless <graph> has only edgeless <graph minors>, and none satisfies $2\delta(H)\geq|H|+4k-1$. We use the intended <dense minor with bounded order and high minimum degree> lemma with the hypothesis $e(G)\geq11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.
\b[Probabilistic upper bound and the constant seven.] Suppose $G$ is nonempty and $e(G)\geq7t\sqrt{\log t}|G|$, and set
$$
k=\left\lfloor\frac7{11}t\sqrt{\log t}\right\rfloor,\qquad
\ell=\lceil\sqrt{\log t}\rceil.
$$
The corrected lemma gives a <graph minor> $H$ of order $N\leq11k+2$ with $2\delta(H)\geq N+4k-1$. Since $\delta(H)\leq N-1$, we also have $N\geq4k+1$. For large $k$, every <vertex> has at most $N/3$ nonneighbours, counting itself, because
$$
N-\delta(H)\leq N/2-2k+1/2\leq N/3;
$$
the last inequality uses $N\leq11k+2\leq12k-3$. Moreover any two distinct <vertices> have at least
$$
2\delta(H)-N\geq4k-1
$$
<common neighbours>.
Choose $2t$ disjoint random $\ell$-sets $A_1,\ldots,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell<4k-1\leq N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of <vertices> having no neighbour in $A$. For each <vertex>, the <probability> that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell}$, even when sampling without replacement. By linearity of <expected value>,
$$
\mathbb E|U(A)|\leq N3^{-\ell}.
$$
Call $A$ good if $|U(A)|\leq N3^{-19\ell/20}$. The <Markov inequality> shows that a random set is bad with <probability> at most $3^{-\ell/20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.
Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no <edge> between them, all its members must lie in $U(A_i)$. Hence
$$
\mathbb P(A_i\text{ good and }e(A_i,A_j)=0)
\leq\left(\frac{N3^{-19\ell/20}}{N-\ell}\right)^\ell.
$$
Here $(N/(N-\ell))^\ell=\exp(O(\ell^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at most
$$
\binom{2t}{2}(1+o(1))3^{-19\ell^2/20}
=O\left(t^{2-(19/20)\log3}\right)=o(t),
$$
since $(19/20)\log3>1$. Applying the <Markov inequality> to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.
Make each selected set connected by fixing one of its <vertices> as a root and, for each of its other members, adding one unused <common neighbour> of that member and the root. This uses at most $t(\ell-1)$ additional <vertices>. Then for each pair with no original <edge>, choose one <vertex> in each set and add an unused <common neighbour> to one of the sets. The added <vertex> is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.
Throughout this process the total number of occupied <vertices> is at most
$$
t\ell+t(\ell-1)+t=2t\ell<4k-1.
$$
Since every pair has at least $4k-1$ <common neighbours>, an unused choice always exists. The final sets are disjoint connected <branch sets of a graph minor> with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of <graph minors>. This <random branch-set construction of a complete graph minor> proves
$$
\boxed{c(t)\leq7t\sqrt{\log t}\quad\text{for sufficiently large }t.}
$$
\b[The sharp linear threshold for a four-vertex complete minor.] We prove by induction that a <graph> on $n\geq4$ <vertices> with at least $2n-2$ <edges> has a $K_4$ <graph minor>. The case $n=4$ is $K_4$ itself. If some <edge> $uv$ has at most one <common neighbour>, an <edge contraction> reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\leq2$ <edges>. The contracted <graph> has at least $2(n-1)-2$ <edges>, so induction applies.
Otherwise every <edge> is in at least two <triangles in a graph>. Choose a <vertex> $v$ incident with an <edge>. Each <vertex> $u\in N(v)$ has at least two <neighbours of a vertex> in $N(v)$, since these are the <common neighbours> of $u,v$. Thus $G[N(v)]$ has <minimum degree of a graph> at least two. A longest path in this finite <graph> has an endpoint adjacent to an earlier nonconsecutive path <vertex>, and hence contains a <cycle in a graph>. Together with $v$, this <cycle in a graph> gives a <wheel graph> as a <subgraph>. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent <branch sets of a graph minor>. Therefore
$$
\boxed{e(G)\geq2n-2\ \Longrightarrow\ G\succ K_4.}
$$
With $2n-3$ <edges> the answer is \b[no]. Take the <join of graphs> of $K_2$ and an <independent set> of $n-2$ <vertices>. It has $1+2(n-2)=2n-3$ <edges>. Among four disjoint putative <branch sets of a graph minor>, at most two can contain the two <vertices> of $K_2$. Any connected branch set avoiding them is a singleton in the <independent set>. At least two such singleton sets would therefore have no <edge> between them. This excludes a $K_4$ <graph minor> for every $n\geq4$ and proves sharpness.