One cannot do much better: let $\mathbb{P}(X = 2/a) = p$ and $\mathbb{P}(X = b) = 1 - p$, then $$ \mathbb{E} X = \frac{2p}{a} + (1 - p) b, \qquad \mathbb{E} X^{-1} = \frac{ap}{2} + \frac{1 - p}{b} . $$ Solving $\mathbb{E} X = 1$ and $\mathbb{E} X^{-1} = 2$ for $p$ and $b$, we get $$ b = \frac{a - 2}{a - 4} > 0, \qquad p = \frac{2 a}{a^2 - 4a + 8} \in [0, 1]$$ whenever $a > 4$, and clearly $p \approx \frac{2}{a}$ as $a \to \infty$.

Let $Y = X + 1/X - 2$. Then $Y \geqslant 0$ and $\mathbb{E} Y = 2 + 1 - 2 = 1$. Thus, $$ \mathbb{P}(Y > y) \leqslant \frac{\mathbb{E} Y}{y} = \frac{1}{y} $$ for every $y > 0$. If we choose $y = a + 1/a - 2$ for some $a > 1$, then $$ \mathbb{P}(X > a) \leqslant \mathbb{P}(X > a) + \mathbb{P}(X < 1/a) = \mathbb{P}(Y > y) \leqslant \frac{1}{y} = \frac{a}{a^2 - 2 a + 1} . $$ The right-hand side is asymptotically equal to $1/a$ (instead of the trivial bound by $2/a$).

One cannot do much better: let $\mathbb{P}(X = a) = p$ and $\mathbb{P}(X = b) = 1 - p$, then $$ \mathbb{E} X = p a + (1 - p) b, \qquad \mathbb{E} (1/X) = \frac{p}{a} + \frac{1 - p}{b} . $$ Solving $\mathbb{E} X = 2$ and $\mathbb{E} (1/X) = 1$ for $p$ and $b$, we get $$ b = \frac{a - 2}{a - 1} > 0, \qquad p = \frac{a}{a^2 - 2a + 2} \in [0, 1]$$ whenever $a > 1$, and clearly $p \sim 1/a$ as $a \to \infty$.