Are rationally connected varieties uniruled? A uniruled variety is the one which admits a dominant map from $X \times \mathbb P^1$.  I think it is true that uniruled varieties are rationally connected.  Is the converse true?  What about low dimensional cases, i.e. surfaces and threefolds?
 A: Yes, rationally connected varieties are uniruled (a fact which follows essentially by definition).
In characteristic zero, one can think about this also by exploiting the following  characterization of uniruled and rationally connected varieties, that  can be found in Debarre's book Higher dimensional algebraic geometry, Chapter 4. 
Let $X$ be a smooth projective variety of dimension $d$, defined over a field $\mathbb{K}$ of characteristic zero.
Recall that a rational curve $f \colon \mathbb{P}^1 \to X$ is called $r$-free if $f^*T_X \otimes \mathscr{O}_{\mathbb{P}^1}(-r)$ is generated by its global sections. By Grothendieck' s Theorem any vector bundle over $\mathbb{P}^1$ splits as a direct sum of line bundles, so $r$-free means $$f^*T_X = \mathscr{O}_{\mathbb{P}^1}(a_1) \oplus \ldots \oplus \mathscr{O}_{\mathbb{P}^1}(a_d),$$
with $a_i \geq r$ for every $i$. Now one has the following 

Proposition. Assume $\textrm{char}(\mathbb{K})=0$. Then $X$ is uniruled if and only through a general point $x \in X$ there is a $0$-free rational curve.  Moreover $X$ is rationally connected if and only if through a general point $x \in X$ there is a $1$-free rational curve.

Remark. The condition $\textrm{char}(\mathbb{K})=0$ is an essential one. For instance, if $\textrm{char}(\mathbb{K})=p>0$ then the Fermat hypersurface of degree $p^r+1$ in $\mathbb{P}^N$, with $N \geq 4$ and $r \geq 1$, is uniruled by lines, none of which are free. Moreover, if $p^r >N$ any such a hypersurface has ample canonical class, then it contains no free rational curves at all.  
A: Let $k$ be an algebraically closed uncountable field of characteristic zero. If $X$ is rationally connected then through two general points of $X$ there is a rational curve. In particular there is a rational curve through a general point of $X$. This means that the evaluation map
$$ev:\mathbb{P}^1\times Mor(\mathbb{P}^1,X)\rightarrow X$$
is dominant. Since $Mor(\mathbb{P}^1,X)$ has at most countably many components and $X$ is irreducible, there exists a component $Y$ of $Mor(\mathbb{P}^1,X)$ such that  
$$ev:\mathbb{P}^1\times Y\rightarrow X$$
is dominant. Therefore $X$ is uniruled. 
