# Proving inequalities using calculus

#### Poly

##### Member
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
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
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
I'll try to prove the result that I've used.

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#### MarkFL

Staff member
You might also think in terms of a Riemann sum definition of the definite integral.

#### Jameson

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

#### Jameson

Staff member
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
Okay, I thought I was meant to approximate the area and show that it goes over $1$.

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

Staff member
Okay, I thought I was meant to approximate the area and show that it goes over $1$.

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 .

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
Sorry, yes, I should have posted the steps to avoid confusion.

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

#### Poly

##### Member
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
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

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.

#### MarkFL

Staff member
...
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.

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#### Deveno

##### Well-known member
MHB Math Scholar
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
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?