The question:
Let $f(x) = ln(x + \sqrt{3 + x^2}) + \frac{1}{2}(x-2)^2$ with no restrictions on domain.
Determine the Lipschitz constant $L$ for the basic statement:
$$| \nabla f(x) - \nabla f(y)| \leq L |x-y|$$
I calculated $$\nabla f(x) = \frac{1 + (x - 2)\sqrt{x^2 + 3}}{\sqrt{x^2+3}} $$
And I know that
$$| \nabla f(x) - \nabla f(y)| = |\frac{1}{\sqrt{3 + x^2}} - \frac{1}{\sqrt{3 + y^2}} + x - y|\leq L |x-y|$$
But I can't come up with any idea of how to determine $L$ (Lipschitz constant) using the basic statement.
Any help is appreciated.