By the classical regularity theory for the Poisson equation, you have 
$$
 \Vert \nabla \psi \Vert_{L^\infty (B_1)} \le C \bigl(\Vert \Delta \psi \Vert_{L^\infty (B_2)} + \Vert \psi \Vert_{L^\infty (B_2)} \bigr).
$$
By scaling this gives 
$$ R\Vert \nabla \psi \Vert_{L^\infty (B_R)} \le C \bigl(R^2 \Vert \Delta \psi \Vert_{L^\infty (B_{2 R})} + \Vert \psi \Vert_{L^\infty (B_{2 R})} \bigr).
$$
By translating the estimate, you obtain thus for every $R > 0$,
$$ R\Vert \nabla \psi \Vert_{L^\infty (\mathbb{R}^n)} \le C \bigl(R^2 \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)} + \Vert \psi \Vert_{L^\infty (\mathbb{R}^n)} \bigr).
$$
By taking $R = \sqrt{\Vert \psi \Vert_{L^\infty (\mathbb{R}^n)} / \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)}}$, we conclude that 
$$
\Vert \nabla \psi \Vert_{L^\infty (\mathbb{R}^n)} \le 2C \Vert \Delta \psi \Vert_{L^\infty (\mathbb{R}^n)}^{1/2} \Vert \psi \Vert_{L^\infty (\mathbb{R}^n)}^{1/2}.
$$