How to prove $\frac{1}{2n+2}<\int_0^{\frac{\pi}{4}}\tan^nx\,dx<\frac{1}{2n}$

How to prove $$\frac{1}{2n+2}<\int_0^{\frac{\pi}{4}}\tan^nx\,dx< \frac{1}{2n}$$ Set $$A_n=\int_0^{\frac{\pi}{4}}\tan^nx\,dx$$, then we have $$A_n+A_{n+2}=\frac{1}{n+1}$$ and we have $$A_{n+2} < A_n$$ ,so we can get $$\frac{1}{2n+2}< \int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n-2}$$ But how to show that$$\int_0^{\frac{\pi}{4}}\tan^nx\,dx < \frac{1}{2n}$$

Change variable to $$t = \tan x$$, we have

$$I_n \stackrel{def}{=}\int_0^{\pi/4} \tan^n x dx = \int_0^1 \frac{t^n}{1+t^2} dt$$

Notice for $$t \in (0,1)$$, we have $$\frac{1 + t^2}{2} < 1$$. This implies

$$I_n > \int_0^1 \frac{t^n}{1+t^2}\cdot\frac{1+t^2}{2} dt = \frac12\int_0^1 t^n dt = \frac{1}{2(n+1)}$$

On the other direction, AM $$\ge GM$$ tell us $$t = \sqrt{1 \cdot t^2} \le \frac{1+t^2}{2}$$ and the inequality is strict when $$t \ne 1$$. This leads to

$$I_n = \int_0^1 \frac{t^{n-1}}{1+t^2} t dt < \int_0^1 \frac{t^{n-1}}{1+t^2}\cdot \frac{1+t^2}{2} dt = \frac12 \int_0^1 t^{n-1} dt = \frac{1}{2n}$$

For $$n=0$$ there is nothing to prove. For $$n=1$$, this is $$\log\sqrt{2}<\frac12$$, which is obviously true.

We have $$\tan^{n+1}x<\frac12(\tan^n x+\tan^{n+2}x)\text{ for }x\in(0,\pi/4) \tag{1}$$ because $$y\in\mathbb{R}^+\mapsto c^y\in\mathbb{R}^+$$ is strictly convex for $$c>0$$. So you get $$A_{n+1}<\frac1{2(n+1)}$$, or equivalently $$A_n<\frac1{2n}$$.