Relation $R$ is reflexive, since $2+a\times a>0, \forall$ real number $a$.

It is symmetric, since $2+ab>0\Rightarrow 2+ba>0.$
However, $R$ is not transitive. 

Consider three real numbers $2,-\frac{1}{6}$ and $-2$ . We have
$\text{2}+2\times \left(-\frac{1}{6}\right)=\frac{5}{3}>0$
and $\text{2}+\left(-\frac{1}{6}\right)\left(-2\right)=\frac{7}{3}>0$
So, $2R\left(-\frac{1}{6}\right)$ and $\left(-\frac{1}{6}\right)R\left(-2\right)$
But, $2\cancel{R}\left(-2\right)$ since $2+2\left(-2\right)=-2\ngtr 0$