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MHajduk
And here you have my solution, a purely arithmetic one:


29 Nov 2015, 18:00 

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 

tthsqe
OK, so how would you prove
Code: a >= b > 0 implies 2 3*(ab) 6*a + b 6/7 1/7  <=   a * b 49*a 7 

30 Nov 2015, 02:01 

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 Rearrange 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)] 

30 Nov 2015, 04:14 

Xorpd!
The next day after I posted that proof I felt bad because if a = b > 0, it follows that (ab)²/(8a) ≤ (a+b)/2√(ab) ≤ (ab)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(x1)+7(1/x^(6/7)1) > 0 because here 6(x1) < 0. It is true, however because given that f'(x) = 6x13+7/x^(6/7), we see that f'(1) = 0 and f"(x) = 66/x^(13/7) = 6(11/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+13b91a^(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
Code: a > b > 0 implies 2 2 3*(ab) 6*a + b 6/7 1/7 3*(ab)  <   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)*(x1) + *(x1) , 2 or 6 + x 1/7 3 13/7 2   x = *t *(x1) . 7 49 If x < 1, then this r.h.s is > 3*(x1)^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
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(x1)+7(1/x^(6/7)1) > 0 because here 6(x1) < 0. 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
A side note:
I am happy for the threadstarter, 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
YONG wrote: A side note: 

04 Dec 2015, 19:23 

MHajduk


13 Dec 2015, 21:08 

Tyler
MHajduk, those visualizations are nice. What did you use to make them?


14 Dec 2015, 15:28 

MHajduk
Tyler wrote: MHajduk, those visualizations are nice. What did you use to make them? 

14 Dec 2015, 19:09 

MHajduk
Another Archimedean figure:


14 Dec 2015, 22:25 

MHajduk


29 Dec 2015, 00:10 

MHajduk


29 Dec 2015, 11:13 

MHajduk


03 Jan 2016, 02:52 

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 nonorientable surface. Thanks!


03 Jan 2016, 11:52 

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 nonorientable surface. Thanks! 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


05 Jan 2016, 01:00 

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