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

Question

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?

Answer 1

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

Answer 2

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 $x \in \mathbb{N}$

Answer 3

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.