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Author
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
And here you have my solution, a purely arithmetic one:

29 Nov 2015, 18:00
tthsqe

Joined: 20 May 2009
Posts: 724
tthsqe
MHajduk, If you don't like calculus, maybe you should post and inequality that can't be tackled by an application of it.
29 Nov 2015, 18:38
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
tthsqe wrote:
MHajduk, If you don't like calculus, maybe you should post and inequality that can't be tackled by an application of it.
I didn't say that I don't like calculus. I have just only shown that this problem may be solved arithmetically, just like that.

BTW, you have answered to my question with some supposition. It's not the thing I have expected.
29 Nov 2015, 20:51
tthsqe

Joined: 20 May 2009
Posts: 724
tthsqe
OK, so how would you prove

Code:
```a >= b > 0 implies

2
3*(a-b)      6*a + b     6/7   1/7
--------- <= --------- - a   * b
49*a           7                     ```
30 Nov 2015, 02:01
YONG

Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
Code:
```Define f(x) = 3x^2 - 13x + 49x^(1/7) - 39.

f(0) = -39 < 0
f(1) = 0

f'(x) = 6x - 13 + 7/x^(6/7) = 6 (x - 1) + 7 [1/x^(6/7) - 1]
Thus, f'(x) > 0 on the interval (0, 1).

Since f(x) is increasing on (0, 1), we have f(x) =< f(1) = 0.

Now, put x = b/a. Note that 0 < x =< 1.

3(b/a)^2 - 13(b/a) + 49(b/a)^(1/7) - 39 =< 0

Multiply both sides by a^2 (>0):

3b^2 - 13ab + 49 a^(13/7) b^(1/7) - 39a^2 =< 0

Re-arrange the terms:

3a^2 - 6ab + 3b^2 =< 42a^2 + 7ab - 49 a^(13/7) b^(1/7)

3 (a^2 - 2ab + b^2) =< 7a (6a + b) - 49a [a^(6/7) b^(1/7)]

3 (a - b)^2 =< 7a (6a + b) - 49a [a^(6/7) b^(1/7)]

3 (a - b)^2 / 49a =< (6a + b)/7 - [a^(6/7) b^(1/7)]    ```
We don't really need partial differentiation for questions like this.

30 Nov 2015, 04:14
Xorpd!

Joined: 21 Dec 2006
Posts: 161
Xorpd!
The next day after I posted that proof I felt bad because if a = b > 0, it follows that (a-b)²/(8a) ≤ (a+b)/2-√(ab) ≤ (a-b)²/(8b) because all of them are zero.
Else if a > b > 0, the way is clear if we multiply through by 2/(√a-√b)².
But tthsqe graciously resurrected my technique of proof by offering a theorem which is much more awkward to solve purely algebraically!
YONG's method of proof is good except that I can't see the insight that allows us to conclude without further discussion that 6(x-1)+7(1/x^(6/7)-1) > 0 because here 6(x-1) < 0.
It is true, however because given that f'(x) = 6x-13+7/x^(6/7), we see that f'(1) = 0 and f"(x) = 6-6/x^(13/7) = 6(1-1/x^(13/7)) < 0 for x ∈ (0,1), so f'(x) > 0 for x ∈ (0,1).
The f'(x) that YONG gets is the same as what would be obtained if 49× the original expression were written as f(a,b) ≥ 0 and then ∂f/∂b were taken.
An alternative might be to multiply by 49a and then we get f(a,a) = 0 and ∂f/∂a = 78a+13b-91a^(6/7)b^(1/7) > 0 for a > b > 0, which we could prove by analyzing ∂²f/∂a² as before or by noting that ∂f/∂a = 91((6a+b)/7-(a⁶b)^(1/7) > 0 for a > b > 0 because the arithmetic mean of a set of positive numbers is greater than or equal to the geometric mean of the set with equality only if all numbers in the set are equal.
30 Nov 2015, 15:25
tthsqe

Joined: 20 May 2009
Posts: 724
tthsqe
Code:
```a > b > 0 implies
2                                2
3*(a-b)    6*a + b    6/7  1/7   3*(a-b)
-------- < ------- - a   *b    < --------
49*a        7                    49*b

Proof: With f(x) = x^(1/7), Taylor's Theorem says
that there is a t between 1 and x such that
f''(t)      2
f(x) = f(1) + f'(1)*(x-1) + ------*(x-1)  ,
2
or
6 + x    1/7   3   -13/7      2
----- - x    = --*t     *(x-1)  .
7            49
If x < 1, then this r.h.s is > 3*(x-1)^2/49.
Setting x = b/a gives the first inequality after homogenizing.

The second inequality follows from
using f(x) = x^(6/7) and setting x = a/b.    ```
01 Dec 2015, 01:22
YONG

Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
Xorpd! wrote:
YONG's method of proof is good except that I can't see the insight that allows us to conclude without further discussion that 6(x-1)+7(1/x^(6/7)-1) > 0 because here 6(x-1) < 0.
I am so sorry because I thought it was an obvious result of YONG's Theorem, which states:
Code:
```Given a positive integer n. The function g(x) > 0 for all x in (0, 1), where
g(x) = n (x - 1) + (n + 1) { 1 / x^[n/(n+1)] - 1 }.

Proof:
g is continuous on (0, 1).
(As x tends to 0+, g(x) tends to positive infinity.)
g(1) = 0.
g'(x) = n { 1 - 1 / x^[(2n+1)/(n+1)] } < 0 for 0 < x < 1.
So, g is decreasing on (0, 1).
Thus, g(x) > g(1) = 0 for 0 < x < 1.    ```
01 Dec 2015, 03:41
YONG

Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
A side note:

I am happy for the thread-starter, MHajduk, that this topic is gaining more and more interest in the forum. It seems that my suggestion to keep everything in one thread is correct after all.

01 Dec 2015, 03:47
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
YONG wrote:
A side note:

I am happy for the thread-starter, MHajduk, that this topic is gaining more and more interest in the forum. It seems that my suggestion to keep everything in one thread is correct after all.

Thank you for your words, I appreciate them.
04 Dec 2015, 19:23
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk

13 Dec 2015, 21:08
Tyler

Joined: 19 Nov 2009
Posts: 1216
Location: NC, USA
Tyler
MHajduk, those visualizations are nice. What did you use to make them?
14 Dec 2015, 15:28
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Tyler wrote:
MHajduk, those visualizations are nice. What did you use to make them?
Blender + GeoGebra + IrfanView + heavily customized LaTeX
14 Dec 2015, 19:09
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Another Archimedean figure:

14 Dec 2015, 22:25
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk

29 Dec 2015, 00:10
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
29 Dec 2015, 11:13
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
03 Jan 2016, 02:52
YONG

Joined: 16 Mar 2005
Posts: 8000
Location: 22° 15' N | 114° 10' E
YONG
Please show us the making of the Klein bottle. Also explain, with the aid of 3D illustrations, why it is an example of a non-orientable surface. Thanks!

03 Jan 2016, 11:52
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
YONG wrote:
Please show us the making of the Klein bottle. Also explain, with the aid of 3D illustrations, why it is an example of a non-orientable surface. Thanks!

OK, I'll try.

Below you have a variation on the theme of sphericon - a hexagonal sphericon (the plane going through two apexes has a shape of the regular hexagon):

03 Jan 2016, 20:16
MHajduk

Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk

05 Jan 2016, 01:00
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