Question

\[\mathsf{\text{Let \(\mathsf{f(x)}\) be a differential and an invertible function, such that \(\mathsf{f''(x)>0}\) and \(\mathsf{f'(x)>0}\).}}\]
\(\mathsf{\text{Prove that}}\)\[\left.\begin{array}{r}&&&&\mathsf{ f^{-1}\left(\frac{x_1 + x_2 +x_3}{3} \right) > \frac{f^{-1}(x_1)+f^{-1}(x_2)+f^{-1}(x_3)}{3}}\end{array}\right.\]

Answer 1

Wow, this is cool!

Answer 2

such that f''<what?
can't read it...

Answer 3

>*

Answer 4

\[\mathsf{\text{I know! :-D}}\]

Hmm it's f"(x) > 0 @turing

Answer 5

I am trying to do it graphically.

Okay!

"Let \(\mathsf{f(x)}\) be a differential and an invertible function, such that \(\mathsf{f''(x)>0}\) and \(\mathsf{f'(x)>0}\).
Prove that\[\left.\begin{array}{r}&&&&\mathsf{ f^{-1}\left(\frac{x_1 + x_2 +x_3}{3} \right) > \frac{f^{-1}(x_1)+f^{-1}(x_2)+f^{-1}(x_3)}{3}}\end{array}\right.\]"

Answer 6

We have a function f, that's increasing and concave up. We can say that its inverse is also increasing, (but concave down).

Answer 7

Not entirely sure about what's between the brackets.

Answer 8

Random co-ordinates

Answer 9

Is this an IMO problem?

Answer 10

No, I don't think so. It's from my textbook

Answer 11

What's the subject? That could probably help.

Answer 12

Invertible functions.

The question has two inequalities to prove.

\[1. \left.\begin{array}{r}&&&&\mathsf{ f^{-1}\left(\frac{x_1 + x_2 +x_3}{3} \right) > \frac{f^{-1}(x_1)+f^{-1}(x_2)+f^{-1}(x_3)}{3}}\end{array}\right.\]
\[2. \left.\begin{array}{r}&&&&\mathsf{ f\left(\frac{x_1 + x_2 +x_3}{3} \right) < \frac{f(x_1)+f(x_2)+f(x_3)}{3}}\end{array}\right.\]

Answer 13

The Second Part is quite clear but I am confused on First part, I am not able to think of it's inverse .

Answer 14

Ok. First
\[ (f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))} \]
Hence if f'(x) > 0 for all x, then
\[ (f^{-1})'(x) > 0 \ \ \ \hbox{ for all } \ x \]
Similarly,
\[ (f^{-1})''(x) > 0 \ \ \ \hbox{ for all } \ x \]

Now given that, you should find it not to hard to prove your #1 above. Call the function g if you like; get away from the complication of f^-1.

Just look at the function and convince yourself that
for any function g such that g' > 0, g'' > 0
it would satisfy that inequality.

Answer 15

correction:
\[ (f^{-1})''(x) < 0 \]
less than. Very important! You should prove that this is the case.

Answer 16

and likewise g'' < 0

Answer 17

Actually, I've already stated that (without proof), but I don't really know how to take it from there.

Answer 18

It follows from the fact the functions are convex. By definition a function is convex if
for all a < b, the line segment connecting (a,f(a)) and (b,f(b)) lies above the graph of f between a and b.

This is equivalent to the statement that for all a, b and x such a < x < b,
\[ \frac{f(x) - f(a)}{x - a } < \frac{ f(b) - f(a)}{b-a} \]

It's a basic result in calculus that if f is differentiable and f' is increasing, then f is convex and not hard to prove using this second definition.

One can also show that if a function f is convex, then for all x, y,
\[ f(tx + (1-t)y) < t f(x) + (1-t) f(y) \ \ \ \hbox{ for } \ 0 < x < 1 --- (*) \]

Let's look at the case
f' > 0 and f'' > 0. This function is convex. Then applying (*) twice
\[ f(t ( u x_1 + (1-u)x_2) + (1-t)x_3) \]
\[< t f(ux_1 + (1-u)x_2) + (1-t)f(x_3) \]
\[ < ut f(x_1) + t(1-u)f(x_2) + (1-t) f(x_3) \]

Choose now t = 2/3, u = 1/2, then
\[ f(\frac{x_1 + x_2 + x_3}{3}) < \frac{ f(x_1) + f(x_2) + f(x_3)}{3} \]

Now use the same procedure \( \it mutatis \ mutandis \) to show that for the function g above
\[ g(\frac{x_1 + x_2 + x_3}{3}) > \frac{ g(x_1) + g(x_2) + g(x_3)}{3} \]

======

For the record, given that
\[ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} \] and f' > 0, f'' > 0, it follows using the chain rule that
\[ (f^{-1})''(x) = -\frac{(f^{-1})'(x)}{f''(f^{-1}(x))} = - \frac{1}{f'(f^{-1}(x)f''(f^{-1}(x))} < 0 \]

Answer 19

Thank you James!

Answer 20

Nicely done :D

Answer 21

Ugh, that last line isn't correct. Same result.
\[ (f^{-1})''(x) = \left( \frac{1}{f'(f^{-1}(x))} \right)' = -\frac{(f^{-1})'(x)f''(f^{-1}(x))}{(f'(f^{-1}(x)))^2} = - \frac{f''(f^{-1}(x))}{(f'(f^{-1}(x)))^3 } < 0\]

Answer 22

[I think I can say that's the first time in my life I've written down the second derivative of the inverse; and now I know why I've not seen it before. Pesky. It is also the first time on OpenStudy I've used 'mutatis mutandis' and I promise not to use it again for a while. ;-) ]

Answer 23

Wow, thanks!

Answer 24

It's clear that this result generalizes. Suppose f is convex on an interval J. For any
\[ x_1, x_2, ..., x_n \in J \]
and constants
\[ 0 < p_1, p_2, ..., p_n < 1 \ \hbox{ such that } \ \sum_ {i=1}^n p_i = 1 \] we have that
\[ f \left( \sum_ {i=1}^n p_ix_i \right) < \sum_ {i=1}^n p_i f(x_i) \]

Answer 25

It's worth pointing out in your problem that f'' > 0 gives you convexity of f and then helps establish the concavity of f^-1.

The f' > 0 is there for that reason.

But it is _also_ there so you know f^-1 exists.

If I were marking your assignment and you didn't mention this last fact, then I might deduct 1/2 mark. ;-)

Answer 26

lol, 1/2 mark! I thought you were tougher than that :P

Answer 27

you're right. 1 1/2 marks. Very sloppy.

Answer 28

Actually the problem tells us that f is invertible. Plus, you know 1 1/2 is relative, it depends on how much the problem weighs.

Answer 29

They tell you f is invertible? Wimps.

Answer 30

Haha!

Answer 31

I needed to google "Wimps" :D

Answer 32

Awesome JamesJ, such a nice solution I couldn't have done it on my own.

How are you so good? does it takes time to be as good as you or maybe a math major?

Answer 33

I used to be an academic mathematician, so that helps. But there's no magic about that, much as I wish it were otherwise. In other words, getting a title or a degree in itself doesn't suddenly make you better. You build up skills over years of work.

Answer 34

Yeah. I'm a math major but nowhere near him! :( I need to work harder I think!

Answer 35

Thanks again. No doubt, you are the best user on openstudy.