# Proof of a particular inequality

I came up with the following approximation

$$\sqrt[4]{\pi}+\frac{2}{1000}\gtrsim\frac{4}{3}$$

I don't know too much about proving an inequality like this algebraically. I was hoping for an extremely rigorous proof of this (I would definitely appreciate names of theorems). I am just starting to self study computational number theory.

I didn't know how to prove this whatsoever. I would think of using a large finite number of iterations on a Taylor series, but I really had no clue how to use that. Thanks for any help.

An similar question type to this is the following: Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$. My wording is a bit odd in this question, so please note that both questions are very similar. (Solving mine algebraically is really the basis, though)

• Oh yeah, could you please prove assuming you were not given a calculator.
– user285523
Nov 24, 2015 at 1:52
• What do you mean by the symbol $\lesssim$?
– ajd
Nov 24, 2015 at 1:53
• Approximately but still greater than is what that symbol means. You also have it backwards :) @AlexanderDunlap
– user285523
Nov 24, 2015 at 1:55
• Oh, boy... Nov 24, 2015 at 2:38

This isn't really by hand, but here is an explanation of what OP found. We can calculate the continued fraction for $\sqrt[4]{\pi}$ as $$[1;3,55,3,1,1,2,3,37,\ldots]$$ We get a good approximation in a continued fraction by stopping just before a "big" number. Hence we get a quite good estimate as $$\sqrt[4]{\pi}\approx [1;3]=\frac{4}{3}$$ We can refine this by taking more terms: $$\sqrt[4]{\pi}\approx [1;3,55]=\frac{221}{166}\approx 1.3313253\ldots$$ This explains why $\sqrt[4]{\pi}+0.002$ is close to $\frac{4}{3}$. To prove the requested bound, we need to go a bit further. The continued fractions alternate as an overestimate, followed by an underestimate. $[1;3]$ is an overestimate (for $\sqrt[4]{\pi}$), $[1;3,55]$ is an underestimate, and so on. Just a few more terms gives us: $$\sqrt[4]{\pi}>[1;3,55,3,1,1]=\frac{1555}{1168}\approx 1.3313356>\frac{4}{3}-0.002$$ Hence $\sqrt[4]{\pi}+0.002>\frac{4}{3}$.

• It looks as if you didn't read the comment below the OPs question (-1)
– user266519
Nov 24, 2015 at 2:13
• Everyone's right. See, I designed this question with the wanting to solve this algebraically. As I didn't specify this very well, I will give a +1
– user285523
Nov 24, 2015 at 2:21
• I have edited the question earlier
– user285523
Nov 24, 2015 at 2:24
• (a) The OP did not include the restriction to hand-calculation in the question itself. (I don't fault the OP, you understand, it's just an observation.) (b) This answer states upfront that it is not by hand; it is not misleading. (c) It is posted in advance of any answer that is by hand. And (d) it contributes to an understanding of the approximation. I think it is therefore worthy of a compensatory upvote. +1 Nov 24, 2015 at 2:42

Maybe it is worth mentioning that this follows from the well-known approximation/inequality $\frac{22}{7} > \pi$ because $(4/3)^4 > \frac{22}{7}$.

• This proves that $\sqrt[4]{\pi}<\frac{4}{3}$, which is not what is asked here. Nov 24, 2015 at 15:21