“The average rate of change will be the same as the instantaneous rate of change somewhere in the interval” under reasonable assumptions, is the gist of the Mean Value Theorem. A pleasant funny review is in the slow-paced 1966 (yes, 1966!) video “The Theorem of the Mean Policeman” . Slightly more technically, if f(x) is a differentiable function on the open interval (a,b) and continuous on the closed interval [a,b], then there is at least one point z in (a,b) such that

f′(z)=[f(b)—f(a)]/(b—a). In our applications below, we prefer to write this last equation as:

f(b)—f(a)=(b—a)f′(z) (MVT).

The purpose of this post is to see the Mean Value Theorem at work in a couple of places where we may not expect to see it. The second example is much more interesting than the first.

**Example 1. (Warm-up example):** *Which is bigger*

*7^(1/3) + 9^(1/3) or 4?*

If you have a calculator or a phone handy, it will probably tell you that the first expression is approximately 3.993. But that’s cheating. Instead consider the function

f(x)=x^(1/3). A straightforward application of the power rule shows that f′(x) is decreasing for x>0. Hence (using MVT twice):

f(9)—f(8)<f(8)—f(7).

But since f(8)=2, this last inequality gives the desired answer.

**Example 2:** *Prove that in any acute triangle with angles A, B and C, that*

*sinA+sinB+sinC>2.*

I have a vague recollection of trying this problem as part of a Putnam training session as an undergraduate student. Recently I saw it again as a post on Stephen Cavadino’s blog.

In fact I recognize Stephen’s clever solution as being the same one presented at my Putnam club decades ago. It’s such a slick solution that it’s definitely worth reading.

Nevertheless, I’m not sure that I could have come up with it. So here’s an alternative, using MVT.

This solution hopefully also highlights the value of chosing good examples to help understand the problem.

**How to solve it:**

I experimented with the conditions of the problem, quickly realizing why the “acute” hypothesis is necessary (otherwise sinA+sinB+sinC can be made arbitrarily close to 0). However, the biggest angle could be a right angle and still have the inequality holding.

*In fact, a right triangle with hypotenuse 1 makes for an extremely interesting special case.* Indeed if C=π/2 (radian measure) then sinC=1, and our required inequality reduces to a special case of the triangle inequality.

Inspired by the simplicity of this example, I decided to compare an arbitrary acute triangle to it. Since we will be comparing different triangles T and T′, we

let f(T) be the function that adds the sines of the three angles of a triangle.

First let’s rank the acute angles; without loss of generality A≤B≤C. Also, let

D=(π/2)—C. Observe that D<A. (This is because D+C+B<π, but A+C+B=π).

So consider a new triangle T′ with angles A—D, B and C+D (the side lengths won’t matter). The new triangle has a right angle (C+D=π/2) and so by our previous discussion f(T′)>2. It remains to show that

f(T) ≥ f(T′). This follows from:

**Lemma:** *Let D>0 and assume that D ≤ A ≤ π/2—D. Then:*

*sin(A—D)+1≤sinA+sin(π/2—D) —D[cosA—cos(π/2—D)]*

Assuming the lemma is true, we have:

sin(A—D)+1≤sinA+sin(π/2—D)

i.e., we imagine replacing two of the summands in f(T′) with the corresponding summands from f(T).

Hence f(T′) ≤ f(T).

To prove the lemma, we use the mean value theorem twice. Once to show:

sin(A)—sin(A—D) ≥ DcosA (exercise!)

and the other time to show:

sin(π/2)—sin(π/2—D) ≤ Dcos(π/2—D).

Combining these last two inequalities yields the lemma.

Notice that in our application of the lemma, we disregarded the term:

—D[cosA—cos(π/2—D)]. This extra term could easily be used to sharpen the inequality.

Another way to sharpen it is to compare our right triangle T′ with another

right triangle T′′ where the angle B in T′ is replaced by an angle approaching π/2.

This means that f(T′′) is approaching 2.

At the beginning, searching for an example where f would approach 2 was part of the process. Realizing how the inequality may fail to hold for obtuse triangles also suggested having a right triangle. So I see this solution as a natural consequence of playing with the conditions of the problem and experimenting with the right examples.

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