Prove that $\ln(1+\frac{1}{x}) < \frac{1}{({x^2 + x})^{1/2}}$ Prove that$$\ln\left(1+\frac{1}{x}\right) < \frac{1}{({x^2 + x})^{1/2}}.$$
I assumed $$g(x) =  \frac{1}{({x^2 + x})^{1/2}} - \ln\left(1+\frac{1}{x}\right)$$ and got it's derivative which is negative, which means, it is a decreasing function. But, it is a function where $\ln(1+\frac{1}{x})$ never exceeds $\frac{1}{({x^2 + x})^{1/2}}$.Graph for the function Also, $g(x)$ don't have any extremum points in $x > 0$. Both tends to $0$ as $x$ tends to $\infty$. How do I prove the inequality?
 A: 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 $\sqrt{1+u} < 1 + \frac u2$.
A: I'll assume you want to show the inequality on $(0,\infty)$. It looks like you want to use the following theorem: Given two continuous functions $f , g \colon [a,\infty) \to \mathbb R$ which are differentiable on $(a , \infty)$ such that $f(a) < g(a) $ and $f'(x) < g'(x)$ for all $x \in (a,\infty)$ we have $f(x) < g(x)$ for all $x \in [a,\infty)$. It sounds like you already computed that $f'(x) < g'(x)$ for all $x \in (0,\infty)$ so we just need points $a_n \in \mathbb R$ with $a_n \to 0$ such that $f(a_n) < g(a_n)$. But for that we calculate $$\frac{1}{\sqrt{\frac{1}{n^2} + \frac 1 n}} > \frac{1}{\sqrt{2 \frac{1}{n^2}}} = \frac{n}{\sqrt 2} > \ln (1+n)$$for any $n \in \mathbb N$.
