= Solution
\b[False: free rank is not monotone under subgroup embeddings.] In fact every finite-rank <free group> embeds in $F_2$, hence in every $F_m$ with $m>1$. Let $a,b$ freely generate a rank-two subgroup and set $z_i=a^iba^{-i}$ for $0\leq i<n$.
To prove that these form a <free basis of a group> for their subgroup, take a nonempty reduced word and combine each maximal run with the same index into a block $z_{i_1}^{e_1}\cdots z_{i_r}^{e_r}$, with $e_j\ne0$ and $i_{j+1}\ne i_j$. Its expansion is
$$
a^{i_1}b^{e_1}a^{i_2-i_1}b^{e_2}\cdots a^{i_r-i_{r-1}}b^{e_r}a^{-i_r}.
$$
Every interior $a$ power and every $b$ power is nonzero, so the free reduction leaves a nonempty word. This proves the <conjugates of one free generator freely generate> property and injectivity of $F_n\to F_m$. In particular,
$$
\boxed{F_3\hookrightarrow F_2,\qquad n>m\text{ is possible}.}
$$
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