Plug $x=\cot(\phi)$ and turn your inequality into
$$ x \ln x < \frac{x^2-1}{2},$$
where $x>1$. The RHS is the second order Taylor approximation of the LHS around $x=1$. Hence the inequality follows from concavity of the derivative of $x \ln x$, which is $1+\ln x$.

This argument has led me to the following elegant solution: Integrate the following well-known inequality between 1 and $x$:

$$ \ln t \le t-1,$$
and get
$$ x \ln x - x + 1 \le \frac{(x-1)^2}{2} \implies $$
$$x\ln x \le \frac{x^2-1}{2}.$$
(Equality iff $x=1$).

[EDIT] Turns out there's a reference for the inequality $\log(1+x) < x-\frac{x^2}{2(1+x)}$, which is equivalent to your inequality.

Hermite–Hadamard inequality states the following: If $f:[a,b]\to \mathbb{R}$ is convex and continuous, then
$$f\left( \frac{a+b}{2}\right) \le \frac{1}{b - a}\int_a^b f(x)\,dx \le \frac{f(a) + f(b)}{2}.$$
The proof is geometric - one compares the integral to a trapezoid and to a rectangle. Applying this to $f(x)=-\frac{1}{x}$ (as you have done below), one obtains the desired inequality.

In fact, this was the **first** application of this inequality. In the letter "Sur deux limites d'une intégrale d´e finie", published by Hermite in Mathesis 3 (1883, p. 82), he proves the above inequality and applies it to $f(x)=\frac{1}{x+1}, a=0$ and gets
$$x −\frac{x^2}{2+x^2} < \log(1+x) < x-\frac{x^2}{2(1+x)}.$$

I was not able to locate the letter myself, but the relevant excerpt appears right in the beginning of the introduction to this monograph by Dragomir and Pearce.