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Proving inequalities using calculus

Poly

Member
Nov 26, 2012
32
This might sound an odd/inappropriate request, but could someone post some inequalities that can be proven using calculus?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
here's one you can try.

prove:

$$\int_1^2 \frac{1}{t}\ dt < 1 < \int_1^3 \frac{1}{t}\ dt$$

to conclude that $2 < e < 3$.

oh, and...no fair using logarithms (pretend you've never heard of them).
 

Poly

Member
Nov 26, 2012
32
here's one you can try.

prove:

$$\int_1^2 \frac{1}{t}\ dt < 1 < \int_1^3 \frac{1}{t}\ dt$$

to conclude that $2 < e < 3$.

oh, and...no fair using logarithms (pretend you've never heard of them).
Is this right for the first part of the inequality? Drawing the graph of $y = \frac{1}{t}$ and $y = 1$ on the interval $t \in [1, 2]$ we see that $\int_{1}^{2}\frac{1}{t} \ dt < \int_{1}^{2}\ dt = 1$ (I posted a diagram but it wasn't rendering well). For the second part, I found the following online:

Jensen's inequality: Let $f(x)$ be a convex function on $[a, b]$. Then $f\left(\frac{a+b}{2}\right) \le \frac{1}{b-a}\int_{a}^{b}f(x)\ dx$.

It also said that in the case where the function strictly convex we have $<$ rather than $\le$.

Let $f(t) = \frac{1}{t}$. Then $\displaystyle f''(t) = \frac{2}{t^3} > 0$ for $t\in\mathbb{R}^+$. Therefore $f(t)$ is strictly convex on $[1, 3]$.

$ \displaystyle \frac{2}{3+1} < \frac{1}{3-1}\int_{1}^{3}\frac{1}{t}\ dt \implies \frac{1}{2} < \frac{1}{2} \int_{1}^{3} \frac{1}{t} \ dt \implies 1 < \int_{1}^{3} \frac{1}{t} \ dt.$ Did I get that right?
 
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Poly

Member
Nov 26, 2012
32
I'll try to prove the result that I've used. (Thinking)
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You might also think in terms of a Riemann sum definition of the definite integral.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
$ \displaystyle \frac{2}{3+1} < \frac{1}{3-1}\int_{1}^{3}f(t)\ dt \implies \frac{1}{2} = \frac{1}{2} \int_{1}^{3} \frac{1}{t} \ dt \implies 1 < \int_{1}^{3} \frac{1}{t} \ dt.$ Did I get that right?
You have a small typo but this part looks right if you replace an "=" with "<" in one place.

$ \displaystyle \frac{2}{3+1} < \frac{1}{3-1}\int_{1}^{3}f(t)\ dt \implies \frac{1}{2} < \frac{1}{2} \int_{1}^{3} \frac{1}{t} \ dt \implies 1 < \int_{1}^{3} \frac{1}{t} \ dt.$
 

Poly

Member
Nov 26, 2012
32
Thanks, guys.

You might also think in terms of a Riemann sum definition of the definite integral.
I get $\displaystyle \int_{1}^{3}\frac{1}{t} \ dt = \lim_{n\to\infty}2\sum_{i=1}^{n}\frac{1}{n+2i}$ I'm not too sure what to do next, though.

My guess is this is greater than $ \displaystyle 2\sum_{i=1}^{7}\frac{1}{7+2i} > 1$ but I really don't know. (Thinking)
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
I think there's a problem with your setup in the above post, Poly. In the final summation you have $n=7$ which is using 7 sub-intervals to approximate the area (assuming the other part is correct, which I don't think it is). You need to find the sum, just as you wrote before, for $n \rightarrow \infty$.

The way to calculate a definite integral using Riemann sums is by the following:

\(\displaystyle \int_{a}^{b}f(x)dx=\lim_{n \rightarrow \infty} \sum_{k=1}^{n}f(x_k) \Delta x\), where \(\displaystyle \Delta x = \frac{b-a}{n}\) and $x_k=a+k\Delta x$.

So for your problem I believe (but am not 100% sure) that the setup is as follows:

For \(\displaystyle f(t)=\frac{1}{t}\), \(\displaystyle \int_{1}^{3} \frac{1}{t}dt=\lim_{n \rightarrow \infty} \sum_{k=1}^{n} f \left( 1+\frac{2k}{n} \right) \left( \frac{2}{n} \right)\)

This is where I'll stop and let someone else confirm. Maybe this is where MarkFL was going with his suggestion.
 

Poly

Member
Nov 26, 2012
32
Okay, I thought I was meant to approximate the area and show that it goes over $1$. (Doh)

I think your set-up and mine are the same since $\displaystyle f(t) = \frac{1}{t}$ therefore $\displaystyle f \left( 1+\frac{2k}{n}\right) = \frac{1}{1+\frac{2k}{n}}.$

I simplified but didn't say so. Sorry about the confusion.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Okay, I thought I was meant to approximate the area and show that it goes over $1$. (Doh)

I think your set-up and mine are the same since $\displaystyle f(t) = \frac{1}{t}$ therefore $\displaystyle f \left( 1+\frac{2k}{n}\right) = \frac{1}{1+\frac{2k}{n}}.$
Hmm, let me continue to simplify that. You might be right!

\(\displaystyle \frac{1}{1+\frac{2k}{n}}=\frac{1}{\frac{n+2k}{n}}=\frac{n}{n+2k}\)

So now we take \(\displaystyle \left( \frac{n}{n+2k} \right) \left( \frac{2}{n} \right)=\frac{2}{n+2k}\)

Ok, it seems you were correct with the set up! My apologies. Since I didn't see your work and the simplified form is hard to see without doing the work, I assumed incorrectly (Blush).

I don't think taking 7 sub-intervals is enough unless you also show that the approximation using 7 sub-intervals is larger than the true area, which opens up another thing to justify!
 

Poly

Member
Nov 26, 2012
32
Sorry, yes, I should have posted the steps to avoid confusion.

You're right I didn't think through my 7 sub intervals guess (Rofl)
 

Poly

Member
Nov 26, 2012
32
I was reading the wiki article on Riemann sums and it says

The left Riemann sum amounts to an overestimation if f is monotonically decreasing on this interval, and an underestimation if it is monotonically increasing.
We're using a left Riemann sum, so our sum can never exceed the true value? If we manually calculate the sum of the first 7 sub intervals (and this is indeed greater than 1 according to wolfram), wouldn't that be enough?
 

Poly

Member
Nov 26, 2012
32
By the way, I enjoyed that question. Thanks guys.

Does anyone know more inequalities that be proven with calculus?

I found two that look like they could use some calculus (Rofl)

1. $x(1+x)^{-1} < \ln(1+x) < x$ where $-1 < x, \ x \ne 0$.

2. $\alpha (x-1) < x^{\alpha}-1 < \alpha x^{\alpha-1}(x-1)$ where $1 < x, \ 1 < \alpha$

Not sure what to differentiate or integrate though. (Thinking)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...
This is where I'll stop and let someone else confirm. Maybe this is where MarkFL was going with his suggestion.
Yes, exactly.

Consider the left sum:

$\displaystyle \int_1^a\frac{1}{t}\,dt=\lim_{n\to\infty}\left[\sum_{k=0}^{n-1}\left(f(t_k)\Delta t \right) \right]$

where:

$\displaystyle \Delta t=\frac{a-1}{n}$

$\displaystyle t_k=1+k\Delta t=1+k\frac{a-1}{n}=\frac{n+(a-1)k}{n}$

and so:

$\displaystyle \int_1^a\frac{1}{t}\,dt=(a-1)\lim_{n\to\infty}\left(\sum_{k=0}^{n-1}\frac{1}{n+(a-1)k} \right)$

However, now that I look at it, this is only useful to show that:

$\displaystyle \int_1^2\frac{1}{t}\,dt<\int_1^3\frac{1}{t}\,dt$

Let's take a look at this from a differential equations perspective:

$\displaystyle x(y)=\int_1^y\frac{1}{t}\,dt$

Differentiate with respect to y:

$\displaystyle \frac{dx}{dy}=\frac{1}{y}$

Inverting both sides, we now have the IVP:

$\displaystyle \frac{dy}{dx}=y$ where $\displaystyle y(0)=1$

Euler's method gives rise to the recursion:

$\displaystyle y_{n+1}=\left(1+\frac{x_n}{n} \right)^n$

and so:

$\displaystyle y=\lim_{n\to\infty}y_{n+1}=e^{x}$

Then, we may write:

$\displaystyle x=\int_1^{e^x}\frac{1}{t}\,dt$

Hence, the inequality becomes:

$\displaystyle \int_1^{2}\frac{1}{t}\,dt<\int_1^{e}\frac{1}{t}\,dt<\int_1^{3}\frac{1}{t}\,dt$

Since $\displaystyle \frac{1}{t}>0$ where $\displaystyle t\in[1,\infty)$ then it follows that:

$\displaystyle 2<e<3$

edit: Now that I review this, I have not shown the inequality is true, I have assumed it to be true. (Worried)
 
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Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
if you are summing over k = 0 to 6, that is a "left-hand sum" (over-estimate).

you are actually using "right-hand sums" (k = 1 to 7), which are under-estimates. this is good, since this means the actual sum (the integral) is larger, which is what you WANT.
 

Poly

Member
Nov 26, 2012
32
if you are summing over k = 0 to 6, that is a "left-hand sum" (over-estimate).

you are actually using "right-hand sums" (k = 1 to 7), which are under-estimates. this is good, since this means the actual sum (the integral) is larger, which is what you WANT.
Oh I see I mixed the two up. Do you have another delicious question perhaps? (Thinking)