# How can I prove the following inequality? $\tan( \frac{1}{\sqrt{(x+1)^3}} ) \leq \tan( \frac{1}{\sqrt{x^3}} )$

How can I prove the following inequality?

With $$x \in\mathbb N$$

$$\tan( \frac{1}{\sqrt{(x+1)^3}} ) \leq \tan( \frac{1}{\sqrt{x^3}} )$$

I did the following :

$$\tan( \frac{1}{\sqrt{(x+1)^3}} ) \leq \tan( \frac{1}{\sqrt{x^3}} )$$ $$\implies \arctan( \tan( \frac{1}{\sqrt{(x+1)^3}} )) \leq \arctan(\tan( \frac{1}{\sqrt{x^3}}))$$ $$\implies \frac{1}{\sqrt{(x+1)^3}} \leq \frac{1}{\sqrt{x^3}}$$ $$\implies \sqrt{(x+1)^3} \leq \sqrt{x^3}$$ $$\implies \sqrt{x^3} \leq \sqrt{(x+1)^3}$$ $$\implies x \leq x+1$$

I know that tangent is not injective, but when working with natural numbers, does the inequality hold?

• knowing tan(x) is an increasing function is enough , when the domain is natural (in this question) , tan takes input only in 0 to π/2 ... can you proceed ? Aug 19, 2020 at 4:10
• By the way your $\Rightarrow$'s should all be $\Leftarrow$'s. Also, your third last line is wrong and should be deleted. Aug 19, 2020 at 4:19

Note that $$f(x)=\tan x$$ is an increasing function for $$x\in (0, \pi/2)$$ $$f(x)=\tan x \implies f'(x)= \sec^2 x>0$$ So for natural numbers $$x$$ $$(1+x)^{3/2} > x^{3/2} \implies \frac{1}{(1+x)^{3/2}} < \frac{1}{x^{3/2}} \implies \tan\frac{1}{(1+x)^{3/2}} < \tan \frac{1}{x^{3/2}}.$$

Edit:

Note that $$\frac{1}{(1+x)^{3/2}},\frac{1}{(x)^{3/2}}<\pi/2, \text{when}, x=1,2,3,4,...$$

• Thanks, corrected now. Aug 19, 2020 at 4:26
• That wasn't correct, there was an error placing "$" sign, I have suggested edits please review Aug 19, 2020 at 4:57 •$tan(\frac{\pi}{4})= 1, tan(\frac{\pi}{2}+\frac{\pi}{4})= -1$Aug 19, 2020 at 5:09 • Here$x$are natural numbers$ 1/(1+x)^{3/2},,1/x^{3/2}$will always be acute angle for$x=1,2,3,4,5,...$less than$\pi/2$. Aug 19, 2020 at 5:15 • Now , it's ok,,, before edit, you said ,$tan$is increasing on$(0,\infty)\$, that's why i devoted. So, (+1) Aug 19, 2020 at 5:43

You can apply a function to both sides of an inequality and preserve the order as long as the function is strictly increasing. In this case arctan is strictly increasing in (0,$$\tan( \frac{1}{\sqrt{(x+1)^3}})$$) for $$x \in \mathbb{N}$$

We know that, $$n^{3/2}$$ would increase as $$n$$ increases and $$n>0$$ this directly implies that the multiplicative inverse would necessarily decrease and close towards zero as $$n$$ approaches $$\infty$$.

Now, when the domain of $$n$$ is restricted to natural numbers, $$n^{-3/2}$$ will only take values in the range $$(0,1]$$.

Also, it is known that $$1$$ radian ≈ $$57.3^\circ$$, so we can say that as $$n$$ increases, the value of $$\tan {n^{-3/2}}$$ decreases towards zero (input of $$\tan$$ reduces gradually.

Hence proved.