Given $ \begin{eqnarray*} F(x) &=& \displaystyle\int_{0}^{x} \dfrac{1}{\sqrt{1 + t^2}} dt = \log(x + \sqrt{x^2 + 1})\\ G(x) &=& \displaystyle\int_{1}^{x} \dfrac{1}{t} dt = \log(x) \end{eqnarray*} $ Prove that $F(x) \geq G(x) $ for all $x \geq 1$.
My initial approach is to integrate them, but my teacher said there is a direct way using the Fundamental Theorem of Calculus, and I have no idea how could I approach this problem? My thought is to use the lower sum and upper sum, but that doesn't seem promising. So could anyone share me some ideas?
Edit
Please ignore the $\log(x + \sqrt{x^2 +1})$ part because it was the result after I integrated $F(x)$.