This inequality comes from scaling. Assume that $f$ satisfies $|D^2f| \leq 1$ on $\mathbb{R}^n$ and $f \geq 0$. It suffices to prove that $$|\nabla f(0)|^2 \leq 2f(0).$$ Equality holds if $f(0) = 0$, so assume that $f(0) > 0$. We may assume that $f(0) = 1$ after taking the rescaling $\tilde{f}(x) = \lambda^{-2}f(\lambda x)$, with $\lambda^2 = f(0)$, since this rescaling preserves the Hessian of $f$, the sign of $f$, and the ratio of interest: $$|\nabla \tilde{f}(0)|^2 = |\nabla f(0)|^2/f(0).$$ In the situation $f(0) = 1$ it is clear that $|\nabla f(0)|^2 \leq 2$, otherwise the Hessian bound would imply that $f < 0$ somewhere (follow a ray in the direction $-\nabla f(0)$).

The situation when $f$ is a nonnegative $C^{1,\,1}$ function on a compact manifold ($\partial \Omega$) is similar.

This inequality comes from scaling. Assume that $f$ satisfies $|D^2f| \leq 1$ on $\mathbb{R}^n$ and $f \geq 0$. It suffices to prove that $$|\nabla f(0)|^2 \leq 2f(0).$$ Equality holds if $f(0) = 0$, so assume that $f(0) > 0$. We may assume that $f(0) = 1$ after taking the rescaling $\tilde{f}(x) = \lambda^{-2}f(\lambda x)$, with $\lambda^2 = f(0)$, since this rescaling preserves the Hessian of $f$, the sign of $f$, and the ratio of interest: $$|\nabla \tilde{f}(0)|^2 = |\nabla f(0)|^2/f(0).$$ In the situation $f(0) = 1$ it is clear that $|\nabla f(0)|^2 \leq 2$, otherwise the Hessian bound would imply that $f < 0$ somewhere (follow a ray in the direction $-\nabla f(0)$).

The situation when $f$ is a nonnegative $C^{1,\,1}$ function on a compact manifold ($\partial \Omega$) is similar. One can e.g. make a nonnegative extension of $f$ to $\mathbb{R}^n$ with comparable $C^{1,1}$ norm and apply the previous reasoning.

This inequality comes from scaling. Assume that $f$ satisfies $|D^2f| \leq 1$ on $\mathbb{R}^n$ and $f \geq 0$. It suffices to prove that $$|\nabla f(0)|^2 \leq 2f(0).$$ Equality holds if $f(0) = 0$, so assume that $f(0) > 0$. We may assume that $f(0) = 1$ after taking the rescaling $\tilde{f}(x) = \lambda^{-2}f(\lambda x)$, with $\lambda^2 = f(0)$, since this rescaling preserves the Hessian of $f$, the sign of $f$, and the ratio of interest: $$|\nabla \tilde{f}(0)|^2 = |\nabla f(0)|^2/f(0).$$ In the situation $f(0) = 1$ it is clear that $|\nabla f(0)|^2 \leq 2$, otherwise the Hessian bound would imply that $f < 0$ somewhere (follow a ray in the direction $-\nabla f(0)$).

The situation when $f$ is a nonnegative $C^{1,\,1}$ function on a compact manifold (the boundary) is similar.

This inequality comes from scaling. Assume that $f$ satisfies $|D^2f| \leq 1$ on $\mathbb{R}^n$ and $f \geq 0$. It suffices to prove that $$|\nabla f(0)|^2 \leq 2f(0).$$ Equality holds if $f(0) = 0$, so assume that $f(0) > 0$. We may assume that $f(0) = 1$ after taking the rescaling $\tilde{f}(x) = \lambda^{-2}f(\lambda x)$, with $\lambda^2 = f(0)$, since this rescaling preserves the Hessian of $f$, the sign of $f$, and the ratio of interest: $$|\nabla \tilde{f}(0)|^2 = |\nabla f(0)|^2/f(0).$$ In the situation $f(0) = 1$ it is clear that $|\nabla f(0)|^2 \leq 2$, otherwise the Hessian bound would imply that $f < 0$ somewhere (follow a ray in the direction $-\nabla f(0)$).

The situation when $f$ is a nonnegative $C^{1,\,1}$ function on a compact manifold ($\partial \Omega$) is similar.