Prove that $\ln(1+\frac{1}{x}) < \frac{1}{({x^2 + x})^{1/2}}$

It becomes a bit simpler if you substitute $x = 1/u$. Then the inequality becomes $$ \log(1+u) < \frac{u}{\sqrt{1+u}} $$ which holds for $u > 0$ because $$ h(u) = \log(1+u) - \frac{u}{\sqrt{1+u}} $$ satisfies $h(0) = 0$ and $h$ is strictly decreasing: $$ h'(u) = \frac{2\sqrt{1+u} -(2+u)}{2(1+u)^{3/2}} < 0 $$ for $u > 0$ because $\sqrt{1+u} < 1 + \frac u2$.