Inequality involving ArcTan How to prove that for $x\in[0, +\infty]$ the following inequality is true: $$\arctan x\geq\frac{3 x}{1+2\sqrt{1+x^2}}?$$
I don't have idea from where to start, so any hint is welcome. Thanks in advance. 

Let's say $x=\tan \theta$, so that:
$$\theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$
$\sqrt{1+\tan^2\theta}=\sec\theta$ because the range of $\arctan x$ always has $\cos \theta > 0$:
$$\theta \geq \frac{3\tan \theta}{1+2\sec\theta}$$
Multiply both the numerator and denominator by $\cos \theta$:
$$\theta \geq \frac{3\sin\theta}{\cos\theta+2}$$
Now, these two functions intersect at $\theta=0$. Also, $\theta$ always has a derivative of $1$, so if we can prove that the derivative of $\frac{3\sin\theta}{\cos\theta+2}$ is always $\leq 1$, we can prove this inequality for $\theta \geq 0$.
$$\frac{d}{d\theta}\frac{3\sin\theta}{\cos\theta+2}=\frac{(\cos\theta+2)\frac{d}{d\theta}(3\sin\theta)-3\sin\theta\frac{d}{d\theta}(\cos\theta+2)}{(\cos\theta+2)^2}$$
$$\frac{d}{d\theta}\frac{3\sin\theta}{\cos\theta+2}=\frac{3\cos^2 \theta+6\cos\theta+3\sin^2\theta}{(\cos\theta+2)^2}=\frac{3+6\cos\theta}{(\cos\theta+2)^2}$$
Now, we need to prove that this is less than or equal to $1$, which is the same as proving:
$$3+6\cos\theta \leq (\cos\theta+2)^2$$
Expand the right side:
$$3+6\cos\theta \leq \cos^2\theta+4\cos\theta+4$$
Subtract both sides by the left side:
$$0 \leq \cos^2\theta-2\cos\theta+1$$
Factor the right side:
$$0 \leq (\cos\theta-1)^2$$
Now, this inequality is obviously true because $0 \leq u^2$ for all $u \in \Bbb{R}$, so we have proven that the derivative of $\frac{3\sin\theta}{\cos\theta+2}$ is always $\leq 1$, so since $\theta$ and that funciton intersect at $\theta=0$, we can conclude that:
$$\theta \geq 0 \implies \theta \geq \frac{3\sin\theta}{\cos\theta+2}$$
$$\theta \geq 0 \implies \theta \geq \frac{3\tan \theta}{1+2\sqrt{1+\tan^2 \theta}}$$
$$x \geq 0 \implies \arctan x \geq \frac{3x}{1+2\sqrt{1+x^2}}$$
