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Vladimir
Member since Jun2203

Aug1703, 11:44 AM (EST) 

"Proof of inequality"

LAST EDITED ON Aug2503 AT 05:46 AM (EST) This problem has been posted as topic #118 some 11/2 year ago but without any reply. The server will not let me reply to this topic. I tried so many ways to prove the inequality for strictly positive reals a,b and c: 64·a·b·c·(a + b + c)^{3} <= 27.(b + c)^{2}·(c + a)^{2}·(a + b)^{2} Can you give me a hint? Thank you. With respectfully regards. Frc  The inequality can be written as {(a + b + c)/3}^{3}/(abc) £ {(a + b)/2}^{2}/(ab)·{(b + c)/2}^{2}/(bc)·{(c + a)/2}^{2}/(ca) We have the 3rd power of the arithmetic and geometric mean ratio for the triple (a, b, c) on the left and a product of the 2nd powers of the arithmetic and geometric mean ratios for the doubles (a, b), (b, c), and (c, a) on the right. However, the inequality between the geometric and arithmetic means for the triple works against the inequalities between the geometric and arithmetic means for the doubles and consequently, these inequalities cannot be used in the proof. We can scale the inequality by the 4th power of a, b, or c to make at least one number equal to one. Let's scale it by c^{4}. Denote x = a/c > 0 and y = b/c > 0. We have to prove xy·{(x + y + 1)/3}^{3} £ {(x + 1)/2}^{2}·{(y + 1)/2}^{2}·{(x + y)/2}^{2} For x = y = 1, the inequality becomes an equality. I will prove that for x ³ 0, y ³ 0, the following function has a minimum (equal to zero) only at x = y = 1 and x = y = 0: F(x, y) = {(x + 1)/2}^{2}·{(y + 1)/2}^{2}·{(x + y)/2}^{2}  xy·{(x + y + 1)/3}^{3} For a minimum, both partial derivatives must vanish: ¶F/¶x = (x + 1)/2·{(y + 1)/2}^{2}·(x + y)/2·(2x + y + 1)/2  y·{(x + y + 1)/3}^{3}  xy·{(x + y + 1)/3}^{2} = 0 ¶F/¶y = {(x + 1)/2}^{2}·(y + 1)/2·(x + y)/2·{2y + x + 1)/2  x·{(x + y + 1)/3}^{3}  xy·{(x + y + 1)/3}^{2} = 0 This is a set of 2 nonlinear equations for 2 unknowns, which we denote ¶F/¶x = P(x,y) = 0 ¶F/¶y = Q(x,y) = 0 This set of 2 equations can be solved by the method of succesive approximation (see the topic #226  Solving the Equations in the College Math conference): x_{k+1} = x_{k}  D_{x}(x_{k}, y_{k})/J(x_{k}, y_{k}) y_{k+1} = y_{k}  D_{y}(x_{k}, y_{k})/J(x_{k}, y_{k}) where J(x,y) is the Jacobian ¶P/¶x, ¶P/¶y ¶Q/¶x, ¶Q/¶y and D_{x}, D_{y} two other determinants for use in the Cramer's rule: P, ¶P/¶y Q, ¶Q/¶y ¶P/¶x, P ¶Q/¶x, Q I got the following 9 different root pairs with various initial approximations (to get all possible roots): (0, 0) (1, 1) (0.597311, 0.597311) (0, 1) (1, 0) (0.377067, 0.377067) (0.602960, 0.602960) (0.501094, 0.017834) (0.017834, 0.501094) Since x ³ 0, y ³ 0, we have to consider only the nonnegative roots (1st, 2nd and, 3rd). We have to determine if these 2 roots correspond to minima, maxima, or neither. For a function f(x) of one variable, we have a minimum if the second derivative f"(x) > 0 and a maximum if f"(x) < 0. Similarly, for a function F(x, y) of 2 variables, if the determinant from the second partial derivatives is greater than zero, we have either a local minimum (if both ¶^{2}F/¶^{2}x > 0 and ¶^{2}F/¶^{2}y > 0) or a local maximum (if both ¶^{2}F/¶^{2}x < 0 and ¶^{2}F/¶^{2}y < 0). If the determinant from the second partial derivatives is less than zero, we have a saddle point. ¶^{2}F/¶^{2}x, ¶^{2}F/¶x¶y ¶^{2}F/¶x¶y, ¶^{2}F/¶^{2}y This determinant is just the Jacobian J(x, y) = ¶(P, Q)/¶(x, y). Substituting the 3 nonnegative roots into the Jacobian, we find J(0, 0) = 0.000943073 > 0 (minimum) ¶^{2}F/¶^{2}x = 0.03125 > 0 ¶^{2}F/¶^{2}y = 0.03125 > 0 J(1, 1) = 0.0833333 > 0 (minimum) ¶^{2}F/¶^{2}x = 0.333333 > 0 ¶^{2}F/¶^{2}y = 0.333333 > 0 J(0.597311, 0.597311) = 0.01464714 < 0 (saddle point) On top of that, F(x, 0) ³ 0 on the xaxis, F(0, y) ³ 0 on the yaxis, and F(0, 0) = 0 at the origin. Therefore, the original function F(x, y) has only 2 minima in the 1st quadrant at (0, 0) and (1, 1). By direct inspection F(0, 0) = F(1, 1) = 0. Therefore, F(x, y) > 0 for any other x > 0, y > 0. Any other attempt of a proof that I tried failed, and I made quite a few. 

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sfwc
Member since Jun1903

Aug1803, 09:25 PM (EST) 

1. "RE: Proof of inequality"
In response to message #0

LAST EDITED ON Aug2003 AT 05:06 PM (EST) Try multiplying it out and applying AM/GM to the whole thing, rather than just the individual parts. It goes a little something like this: I denote addition by @ due to my problems with getting a + sign let S(l,m,n) = a^l*b^m*c^n @ a^m*b^n*c^l @ a^n*b^l*c^m by AM/GM: a^4*b*c =< (a^4*b^2 @ a^4*c^2)/2 => 2S(4,1,1) <= S(4,0,2) @ S(4,2,0) <1> a^3*b^2*c =< (a^4*c^2 @ b^4*a^2)/2 =>S(3,2,1) =< S(4,0,2) <2> a^3*b^2*c =< (a^3*c^3 @ a^3*b^3 @ a^3*b^3)/3 => S(3,2,1) =< S(3,3,0) <3> a^2*b^2*c^2 =< (a^3*b^3 @ b^3*c^3 @ c^3*a^3)/3 => S(2,2,2) =< S(3,3,0) <4> by symmetry on <2> and <3>: S(3,1,2) =< S(4,2,0) <5> S(3,1,2) =< S(3,3,0) <6> now take 5*<1> @ 22*<2> @ 8*<3> @ 38*<4> @ 22*<5> @ 8*<6>: 10S(4, 1, 1) @ 30S(3, 2, 1) @ 38S(2,2,2) @ 30S(3,1,2) =< 27S(4,0,2) @ 27S(4,2,0) @ 54S(3,3,0) => 64S(4,1,1) @ 192S(3,2,1) @ 192S(3,1,2) @ 128S(2,2,2) =< 27S(4,2,0) @ 27S(4,0,2) @ 54S(4,1,1) @ 54S(3,3,0) @ 162S(3,2,1) @ 162S(3,1,2) @ 90S(2,2,2) now factorising both sides yields the desired result.Thankyou sfwc <>< 

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Vladimir
Member since Jun2203

Aug1903, 01:38 AM (EST) 

2. "RE: Proof of inequality"
In response to message #1

LAST EDITED ON Aug2503 AT 05:43 AM (EST) I tried to multiply it through several times, it was a rather ugly exercise, and it did not seem to lead anywhere.As for the 9 different root pairs: I have 4 local minima, 1 local maximum, and 4 saddle points of the function F(x, y). All the 4 minima are equal to zero, the single maximum is greater than zero, and thee 4 saddle points are also greater than zero: 1. minimum: F(0, 0) = 0 J(0, 0) 0.000943073 > 0 ¶^{2}F/¶^{2}x = ¶^{2}F/¶^{2}y = 0.03125 > 0 2. minimum: F(1, 1) = 0 J(1, 1) = 0.0833333 > 0 ¶^{2}F/¶^{2}x = ¶^{2}F/¶^{2}y = 0.333333 > 0 3. minimum: F(1, 0) = 0 J(1, 0) = 0 ¶^{2}F/¶^{2}x = 0.03125 > 0 ¶^{2}F/¶^{2}y = 0 4. minimum: F(0, 1) = 0 J(0, 1) = 0 ¶^{2}F/¶^{2}x = 0 ¶^{2}F/¶^{2}y = 0.03125 > 0 1. maximum: F(0.377067, 0.377067) = 0.00125982 > 0 J(0.377067, 0.377067) = 0.000135991 > 0 ¶^{2}F/¶^{2}x = ¶^{2}F/¶^{2}y = 0.0138876 < 0 1. saddle F(0.597311, 0.597311) = 0.00548305 > 0 J(0.597311, 0.597311) = 0.0146471 < 0 2. saddle F(0.602960, 0.602960) = 0.0.000682242 > 0 J(0.602960, 0.602960) =0.000367897 < 0 3. saddle F(0.501094, 0.017834) = 0.00097343 > 0 J((0.501094, 0.017834) = 0.000290349 < 0 4. saddle F(0.017834, 0.501094) = 0.00548305 > 0 J(0.017834, 0.501094) = 0.0146471 < 0 I can therefore claim that F(x, y) ³ 0 everywhere, not just for x > 0 and y > 0. I got this inequality by dividing the original inequality for a, b, c by c^{4} > 0 (i.e., assuming c ¹ 0). By direct inspection, the original inequality becomes an equality for a = b = c = 0. Therefore, I can claim that the original inequality for a, b, c holds for any real a, b, c, not just for a > 0, b > 0, c > 0. The only problem I have with my proof is that I did not prove (and do not know how to prove) that I got all the possible root pairs of the 2 equations P(x, y) = 0 and Q(x, y) = 0 and consequently, all the possible local extrema of the function F(x, y). I made a 3D plot of this function and it clearly does not have any more extrema, but how to prove it? 

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Vladimir
Member since Jun2203

Aug2203, 06:02 AM (EST) 

3. "RE: Proof of inequality"
In response to message #2

LAST EDITED ON Aug2603 AT 11:06 PM (EST) I already know a lot about the roots. First, let's look for all roots on the line y = x.P(x, y) = (x + 1)/2·{(y + 1)/2}^{2}·(x + y)/2·(2x + y + 1)  y·(4x + y + 1)/3·{(x + y + 1)/3}^{2} = 0 Q(x, y) = (y + 1)/2·{(x + 1)/2}^{2}·(x + y)/2·(2y + x + 1)  x·(4y + x + 1)/3·{(x + y + 1)/3}^{2} = 0 Since P(x, y) = Q(y, x), it does not matter whether I substitue y = x to P(x, y) = 0 or Q(x, y) = 0: P(x, x) = {(x + 1)/2}^{3}·x·(3x + 1)  x·(5x + 1)·{(2x + 1)/3}^{2} = 0 The first root is x = 0. Factoring it out yields 27·(x + 1)^{3}·(3x + 1)  16·(2x + 1)^{2}·(5x + 1) = 81x^{4}  50x^{3}  60x^{2} + 11 = 0 I already know that x = 1 is a root. Factoring it out yields 81x^{3} + 31x^{2}  29x  11 = 0 Finding roots of this 3rd order polynomial by the method of successive approximations: x_{k+1} = x_{k}  (81x_{k}^{3} + 31x_{k}^{2}  29x_{k}  11)/(3·81x_{k}^{2} + 2·31x_{k}  29) I get 3 different real roots: x = 0.597311 x = 0.377067 x = 0.602960 Together with x = 0 and x = 1, these are just the 5 roots I got before, for which the xroot was the same as the yroot. There are no other roots on the line y = x. To find the remaining roots, I can assume x ¹ y. Next, I make 2 combinations of P(x, y) and Q(x, y) to cancel out any identical factors (but not factors symmetrical with respect to x and y): P(x, y)  Q(x, y) = 0 x·(4y + x + 1)·P(x, y)  y·(4x + y + 1)·Q(x, y) = 0 Using the following identit's: (y + 1)·(2x + y + 1) = (x + y + 1)^{2}  x^{2} (x + 1)·(2y + x + 1) = (x + y + 1)^{2}  y^{2} I simplify the 1st equation to: 1/32·(x + 1)·(y + 1)·(x + y)·{(x + y + 1)^{2}  x^{2}}  1/32·(x + 1)·(y + 1)·(x + y)·{(x + y + 1)^{2}  y^{2}} = = 1/27·(x + y + 1)^{2}·(y^{2}  x^{2} + y  x) 1/32·(x + 1)·(y + 1)·(x + y)·(y^{2}  x^{2}) = 1/27·(x + y + 1)^{3}·(y  x) and factoring out (x  y) yields 27·(x + 1)·(y + 1)·(x + y)^{2} = 32·(x + y + 1)^{3} Then I simplify the 2nd equation to: {(x + y + 1)^{2}  x^{2}}·x·(4y + x + 1) = {(x + y + 1)^{2}  y^{2}}·y·(4x + y + 1) (x + y + 1)^{3}·(x  y) = 4xy·(x^{2}  y^{2}) + x^{4}  y^{4} + x^{3}  y^{3} and factoring out (x  y) yields (x + y + 1)^{3} = 2xy·(x + y) + (x + y)^{3} + (x + y)^{2}  xy 2·(x + y)^{2} + 3·(x + y) + 1 = xy·{2·(x + y)  1} Now I make a substitution u = x + y v = x  y (x + 1)·(y + 1) = {(u + 2)^{2}  v^{2}}/4 xy = (u^{2}  v^{2})/4 From the 1st equation I get 27·{(u + 2)^{2}  v^{2})·u^{2}  128·(u + 1)^{3} v^{2} = (u + 2)^{2}  128/27·(u + 1)^{3}/u^{2} and from the 2nd equation I get 2u^{2} + 3u + 1 = (u^{2}  v^{2})/4·(2u  1) v^{2} = u^{2}  4·(2u^{2} + 3u + 1)/(2u  1) Eliminating v^{2} from these 2 equations I get a 4th order polynomial exclusively in u: u^{2}  4·(2u^{2} + 3u + 1)/(2u  1) = (u + 2)^{2}  128/27·(u + 1)^{3}/u^{2} 4·(u + 1)·(2u  1)·u^{2}  4·(2u^{2} + 3u + 1)·u^{2} = 128/27·(u + 1)^{3}·(2u  1) I already know that the pairs (x, y) = (0, 1) and (x, y) = (1, 0) are the roots of P(x,y) and Q(x, y) equations. In both cases, u = x + y = 1. The last equation must have this root. Consequently, the quadratic polynomial 2u^{2} + 3u + 1 must be factorable by (u + 1): 2u^{2} + 3u + 1 = (u + 1)·(2u + 1) Factoring (u + 1) from the equation yields (2u  1)·u^{2}  (2u + 1)·u^{2} = 32/27·(u + 1)^{2}·(2u  1) 27u^{3} = 8·(2u^{3} + 3u^{2}  1) 11u^{3}  24u^{2} + 8 = 0 Finding roots of this 3rd order polynomial by the method of successive approximations u_{k+1} = u_{k}  (11u_{k}^{3}  24u_{k}^{2} + 8)/(3·11u_{k}^{2}  2·24u_{k}) I get 3 different real roots: u = 2 u = 0.700746 u = 0.518928 Seeing that the 1st root is an integer, I can factor it out and solve the remaining quadratic equation: 11u^{3}  24u^{3} + 8 = (u  2)·(11u^{2}  2u  4) = 0 u = (1 ± 3Ö5)/11 The other variable v is real only for the last root u = (1  3Ö5)/11 = 0.518928: v^{2} = u^{2}  4·(2u^{2} + 3u + 1)/(2u  1) = (u + 2)^{2}  128/27·(u + 1)^{3}/u^{2} v = Ö(16) = ± 4i v = Ö(40.2005) = ± 6.340385 i v = Ö(0.233540) = ± 0.483260 The other root for which x ¹ y was u = 1, implying v = ± 1. For real x = (u + v)/2 and y = (u  v)/2, I can use only 2 of the uroots: (u, v) = (0.518928, 0.483260) (u, v) = (0.518928, +0.483260) (u, v) = (1, +1) (u, v) = (1, 1) (x, y) = (0.50109, 0.0178338) (x, y) = (0.0178338, 0.50109) (x, y) = (0, 1) (x, y) = (1, 0) All root pairs of the partial derivatives P(x, y) and Q(x, y) of the original function F(x, y) have been accounted for. There are no more real roots of P(x, y) and Q(x, y), and consequently, no more local extrema of F(x, y). The original inequality holds for all real a, b, c, not just for a > 0, b > 0, c > 0. ··· 

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Vladimir
Member since Jun2203

Aug2203, 06:14 AM (EST) 

4. "RE: Proof of inequality"
In response to message #1

LAST EDITED ON Sep0103 AT 09:07 PM (EST) Hi,I am really impressed how fast you figured it out. As for the disappearing "+" characters, check the last message in topic #348  Integration  in the College Math conference. Regards, Vladimir  Your proof in a more readable way: Let S(k,m,n) = a^{k}b^{m}c^{n} + a^{m}b^{n}c^{k} + a^{n}b^{k}c^{m} By AM/GM: a^{4}bc £ (a^{4}b^{2} + a^{4}c^{2})/2 => 2·S(4,1,1) £ S(4,0,2) + S(4,2,0) <1> a^{3}b^{2}c £ (a^{4}c^{2} + b^{4}a^{2})/2 => S(3,2,1) £ S(4,0,2) <2> a^{3}b^{2}c £ (a^{3}c^{3} + a^{3}b^{3} + a^{3}b^{3})/3 => S(3,2,1) £ S(3,3,0) <3> a^{2}b^{2}c^{2} £ (a^{3}b^{3} + b^{3}c^{3} + c^{3}a^{3})/3 => S(2,2,2) £ S(3,3,0) <4> By symmetry on <2> and <3>: S(3,1,2) £ S(4,2,0) <5> S(3,1,2) £ S(3,3,0) <6> Now take 5·<1> + 22·<2> + 8·<3> + 38·<4> + 22·<5> + 8·<6>: 10·S(4,1,1) + 30·S(3,2,1) + 38·S(2,2,2) + 30·S(3,1,2) £ 27·S(4,0,2) + 27·S(4,2,0) + 54·S(3,3,0) => 64·S(4,1,1) + 192·S(3,2,1) + 192·S(3,1,2) + 128·S(2,2,2) £ £ 27·S(4,2,0) + 27·S(4,0,2) + 54·S(4,1,1) + 54·S(3,3,0) + 162·S(3,2,1) + 162·S(3,1,2) + 90·S(2,2,2) Now factorising both sides yields the desired result. 

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Sumudu
guest

Aug2403, 08:18 PM (EST) 

5. "RE: Proof of inequality"
In response to message #4

Here's a different way to do it (I think it's pretty good, there are probably some areas that could be improved though): Let X = a + b, Y = a + c, and Z = b + c a,b,c > 0 implies X + Y > Z, X + Z > Y, and Y + Z > X Therefore, there exists a triangle with sides of length X, Y, and Z. Let x, y, and z represent the angles opposite these sides. Since x + y + z = 180°, cos(z) = cos(x+y) Let A = (x + y)/2, B = (x  y)/2 cos^{2}(A)  cos(A)cos(B) + 1/4 >= cos^{2}(A)  cos(A)cos(B) + cos^{2}(B)/4 = (cos(A)  cos(B)/2)^{2} >= 0 So cos(A)cos(B)  cos^{2}(A) <= 1/4 2cos(A)cos(B)  2cos^{2}(A) <= 1/2 2cos(A)cos(B)  cos^{2}(A) + sin^{2}(A) <= 3/2 (cos(A)cos(B)  sin(A)sin(B)) + (cos(A)cos(B) + sin(A)sin(B))  (cos^{2}(A)  sin^{2}(A)) <= 3/2 cos(A+B) + cos(AB)  cos(2A) <= 3/2 cos(x) + cos(y)  cos(x+y) <= 3/2 cos(x) + cos(y) + cos(z) <= 3/2 3 + cos(x) + cos(y) + cos(z) <= 9/2 (3 + cos(x) + cos(y) + cos(z))/3 <= 3/2 ((1 + cos(x)) + (1 + cos(y)) + (1 + cos(z)))/3 <= 3/2 By AMGM: (1 + cos(x))(1 + cos(y))(1 + cos(z)) <= 27/8 2YZ(1 + cos(x))2XZ(1 + cos(y))2XY(1 + cos(z)) <= 27X^{2}Y^{2}Z^{2} By the Cosine Law: X^{2} = Y^{2} + Z^{2}  2YZcos(x) So that (Y+Z)^{2}  X^{2} = 2YZ(1 + cos(x)) The left hand side is a difference of squares, factoring gives 2YZ(1 + cos(x)) = (X + Y + Z)(X + Y + Z) = 4c(a + b + c) Similarly: 2XZ(1 + cos(y)) = 4b(a + b + c) and 2XY(1 + cos(z)) = 4a(a + b + c) Returning to the inequality, substitution back to a,b,c gives:
64abc(a+b+c)^{3} <= 27(a+b)^{2}(a+c)^{2}(b+c)^{2}, which is what we wanted to prove. The main part that I think needs work is proving cos(x) + cos(y) + cos(z) <= 3/2 (with x + y + z = 180°)  there must be a neater way of doing it. Sumudu 

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Vladimir
Member since Jun2203

Aug2503, 05:49 AM (EST) 

8. "RE: Proof of inequality"
In response to message #5

LAST EDITED ON Aug2503 AT 05:54 PM (EST) Hello Sumudu,Your proof is the best because it is natural. Suppose 0 £ x £ p, 0 £ y £ p, 0 £ z £ p, and x + y + z = p. Then 1 £ cos(x) + cos(y) + cos(z) £ 3/2 Proof:  Since z = p  (x + y), cos(z) = cos(x + y). Define function F(x, y) as F(x, y) = cos(x) + cos(y)  cos(x + y) and let's look for the maximum. Both partial derivatives must vanish: ¶F/¶x = sin(x) + sin(x + y) = 0 ¶F/¶y = sin(y) + sin(x + y) = 0 sin(x) = sin(x + y) = sin(y) Since 0 £ x £ p, 0 £ y £ p, and 0 £ x + y £ p, then either x = y or x = p  y. In the first case, the 2 possibilities are: 1. x = y ¹ 0, z ¹ 0 2. x = y = 0, z = p Only the 1st possibility is acceptable for a triangle with angles x, y, z. Then by symmetry (i.e., by cyclic permutation) of x, y, and z, it must be x = y = z. Since x + y + z = p, 3x = p, x = y = z = p/3. In the second case, the 2 possibilities are: 3. x = 0, y = p, z = 0 4. x = p, y = 0, z = 0 Neither of the last 2 possibilities is acceptable for a triangle with angles x, y, z. The second partial derivatives of function F(x, y) are ¶^{2}F/¶^{2}x = cos(x) + cos(x + y) ¶^{2}F/¶^{2}y = cos(y) + cos(x + y) and determinant D(x, y) of the second partial derivatives is ¶^{2}F/¶^{2}x, ¶^{2}F/¶x¶y = ¶^{2}F/¶x¶y, ¶^{2}F/¶^{2}y = cos(x)·cos(y)  {cos(x) + cos(y)}·cos(x + y) When D(x, y) > 0, we have a local extremum (minimum for ¶^{2}F/¶^{2}x > 0 and ¶^{2}F/¶^{2}y > 0, maximum for ¶^{2}F/¶^{2}x < 0 and ¶^{2}F/¶^{2}y < 0). When D(x, y) < 0, we have a saddle point. 1. maximum x = y = p/3 F(p/3, p/3) = 1/2 + 1/2  (1/2) = 3/2 D(p/3, p/3) = 1/2·1/2  (1/2 + 1/2)·(1/2) = 3/4 > 0 ¶^{2}F/¶^{2}x (p/3, p/3) = 1/2 + (1/2) = 1 < 0 ¶^{2}F/¶^{2}y (p/3, p/3) = 1/2 + (1/2) = 1 < 0 2. saddle point x = y = 0 F(0, 0) = 1 + 1  1 = 1 D(0, 0) = 1·1  (1 + 1)·1 = 1 < 0 3. saddle point x = 0, y = p F(0, p) = 1 + (1)  (1) = 1 D(0, p) = 1·(1)  {1 + (1)}·(1) = 1 < 0 4. saddle point x = p, y = 0 F(p, 0) = (1) + 1  (1) = 1 D(p, 0) = (1)·1  {(1) + 1}·(1) = 1 < 0 We have a single local maximum in the coordinate triangle <(0, 0), (p, 0), (0, p)> at x = y = p/3. Value of this maximum is F(p/3, p/3) = 3/2. On the triangle boundary, F(x, y) = 1: 1. xaxis x Î <0, p>, y = 0 F(x, 0) = cos(x) + cos(0)  cos(x + 0) = cos(0) = 1 2. yaxis x = 0, y Î <0, p> F(0, y) = cos(0) + cos(y)  cos(0 + y) = cos(0) = 1 3. line x + y = p y = p  x, cos(y) = cos(x) F(x, p  x) = cos(x)  cos(x)  cos(p) = cos(p) = (1) = 1 Therefore, 1 £ F(x, y) £ 3/2 for any x, y in the coordinate triangle <(0, 0), (p, 0), (0, p)>. 

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sfwc
Member since Jun1903

Aug3003, 06:18 AM (EST) 

10. "RE: Proof of inequality"
In response to message #8

Thanks, Vladimir. You have restored my faith in the + sign. It is possible to prove cos x + cos y + cos z with a little less algebra, but it unfortunately involves casesplitting: case 1: one of x, y and z, let us say x, is > 2 pi/3 Then cos x + cos y + cos z < 1/2 + 1 + 1 = 3/2 case 2: all of x, y and z are <= 2 pi/3 on this range, the function cos is lower than it's tangent at pi/3, which has gradient g, say. (since cos is concave on <0, pi/2> the gradient is greater then g on the range <0, pi/3> and less than g on the range <pi/3, pi/2>. By symmetry, the gradient at pi  x is equal to that at x for each x, so in particular it is less than g on <pi/2, 2 pi/3>. Combining these gives the result. Although g is easily calculated in this case, I have left it uncalculated for the sake of generality) That is, cos x <= 1/2 + g*(x  pi/3) with similar results for y and z. We deduce: cos x + cos y + cos z =< 3/2 + g*(x + y + z  pi) <=> cos x + cos y + cos z =< 3/2 since x + y + z = pi Thankyou sfwc <>< 

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Sumudu
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Aug2503, 08:10 AM (EST) 

9. "RE: Proof of inequality"
In response to message #7

Oh ok, now what you wrote in the "Integration" topic makes sense to me. I suppose I do prefer my method of proving the cos inequality, I was considering doing it by calculus but opted to find an elementary (so to speak) solution instead. Took me far too long though ... 

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Vladimir
Member since Jun2203

Aug3103, 11:52 PM (EST) 

13. "RE: Proof of inequality"
In response to message #9

Perhaps if you reverse your proof, it looks more natural. Make a substitution a = (x + y)/2 b = (x  y)/2 cos(x) + cos(y)  cos(x + y) = = cos(a + b) + cos(a  b)  cos(2a) = = 2·cos(a)·cos(b)  (2·cos^{2}(a)  1) = = 2·{cos(a)·cos(b)  cos^{2}(a)} + 1 = = 2·{1/4·cos^{2}(b)  (1/2·cos(b)  cos(a))^{2}} + 1 £ £ 2·(1/4·1  0) + 1 = 3/2 The substitution requires some ingenuity, but the rest is routine and natural. Moreover, it is immediately clear that the requirements for equality are cos(b) = 1, b = 0, x = y and cos(a) = 1/2, a = p/3, x + y = 2p/3. Consequently, x = y = p/3 and z = p  (x + y) = p/3.


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Joe
guest

Aug3103, 05:21 PM (EST) 

11. "RE: Proof of Inequality"
In response to message #0

Problem: Let a,b,c>0. Prove 27*(b+c)^2*(c+a)^2*(a+b)^2>=64*abc*(a+b+c)^3. Proof: Since the inequality is homogenous, we may assume without loss of generality that a+b+c=1. 27*(b+c)^2*(c+a)^2*(a+b)^2>=64*abc*(a+b+c)^3 <==>(1a)^2*(1b)^2*(1c)^2>=64/27*abc <==>(1abc+bc+ca+ababc)^2>=64/27*abc <==>(bc+ca+ababc)^2>=64/27*abc <==>(1/a+1/b+1/c1)sqrt(abc)>=8sqrt(3)/9 Smoothing will prove this inequality. If a=b=c=1/3, then the inequality holds. If a,b,c are not all equal, then without loss of generality, let a<1/3<b. Let a'=1/3 and b'=ab/a', then 1/a+1/b1/a'1/b'=(a+ba'b')/(ab). Since a<a'<b'<b and ab=a'b', (a+ba'b')/(ab)>0. Hence (1/a+1/b+1/c1)sqrt(abc) is decreased. Repeating this argument with b and c, we decrease it once again. Now we have a=b=c, which makes the inequality true. 

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Vladimir
Member since Jun2203

Sep0103, 06:11 PM (EST) 

14. "RE: Proof of Inequality"
In response to message #11

LAST EDITED ON Sep0103 AT 11:08 PM (EST)
>(bc+ca+ababc)^2>=64/27*abc ><==>(1/a+1/b+1/c1)sqrt(abc)>=8sqrt(3)/9
You can squareroot an inequality, if the the larger side (after squarerooting) is greater than or equal to zero. You should have pointed out here that 1/a > 1, 1/b > 1, 1/c > 1 as a result of a + b + c = 1 and a > 0, b > 0, c > 0. Therefore, 1/a + 1/b + 1/c  1 > 0 and squarerooting is allowed.
>Since a<a'<b'<b and ab=a'b', (a+ba'b')/(ab)>0.
Too fast. Since b > b' and a < a', b  a > b'  a'.(a + b  a'  b')/(ab) = (a  2·sqrt(ab) + b  a' + 2·sqrt(a'b')  b')/(ab) = {(sqrt(b)  sqrt(a))^2  (sqrt(b')  sqrt(a'))^2}/(ab) > 0
>Hence (1/a+1/b+1/c1)sqrt(abc) is decreased.
(1/a + 1/b + 1/c  1)·sqrt(abc) > (1/a' + 1/b' + 1/c  1)·sqrt(a'b'c)
>Repeating this argument with b and c, we decrease it once again. Now we have a=b=c, ...
How? Now you have a' = 1/3, 1/3 < b' < 1, and 0 < c = 1  (a + b) < 2/3. It can be c < 1/3, c = 1/3, or c > 1/3. I do not see it. 

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Joe
guest

Sep0303, 04:46 PM (EST) 

16. "RE: Proof of Inequality"
In response to message #14

Ok. I should have mentioned that both sides are positive. Sorry. As for the smoothing argument, I guess mine doesn't work, and I don't see a fix right now. But I guess if I had to continue the proof of (1/a 1/b 1/c1)sqrt(abc)>=8sqrt(3)/9, I would let c=1ab and turn it into a two variable inequality. Then proceed from there. Or you could substitute that right from the beginning. 

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Joe
guest

Aug3103, 05:21 PM (EST) 

12. "RE: Proof of inequality"
In response to message #0

Another proof is to expand everything out (which is actually a very feasible approach if you are taking the Olympiad, since they are usually several hours long). This method along with AMGM usually works for all homogenous polynomial inequalities, although sometimes the Schur inequality may be required also. For more information, see the following guide on inequalities written by one of the USA IMO coaches https://www.unl.edu/amc/aactivities/a4forstudents/ineqs080299.ps. You will need both GSView and PostScript to view the PS file. Proofs by expanding everything out is very tedious but almost always will work. Denote by cyc(f) the sum f(a,b,c)+f(b,c,a)+f(c,a,b). 27*(b+c)^2*(c+a)^2*(a+b)^2>=64*abc*(a+b+c)^3 <==>27*(b^2+2bc+c^2)*(c^2+2ca+a^2)*(a^2+2ab+b^2)>=64*abc*(a^3+b^3+c^3+3a^2*b+3a^2*c+3b^2*c+3b^2*a+3c^2*a+3c^2*b+6abc) <==>27cyc(a^4b^2+a^4c^2+2a^3b^3)>=114a^2b^2c^2+10abc*cyc(a^3)+30abc*cyc(a^2b+b^2a) (***) By AMGM the following inequalities hold: 5(a^4b^2+a^4c^2)>=10a^4bc 6(a^4c^2+a^3c^3+b^4a^2+b^4a^2+a^4b^2)>=30a^3b^2c 6(a^4b^2+a^3b^3+c^4a^2+c^4a^2+a^4c^2)>=30a^3c^2b Summing the 3 cyclic permutations of each of the above three, we get 21cyc(a^4b^2+a^2b^4)+12cyc(a^3b^3)>=10abc*cyc(a^3)+30abc*cyc(a^2b+b^2a) (*) Also by AMGM, 6cyc(a^4b^2+a^2b^4)+42cyc(a^3b^3)>=114a^2b^2c^2 (**) Adding (*) and (**) gives the desired (***). 

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Vladimir
Member since Jun2203

Sep0103, 09:12 PM (EST) 

15. "RE: Proof of inequality"
In response to message #12

>Denote by cyc(f) the sum f(a,b,c)+f(b,c,a)+f(c,a,b). >27*(b+c)^2*(c+a)^2*(a+b)^2>=64*abc*(a+b+c)^3 ><==>27*(b^2+2bc+c^2)*(c^2+2ca+a^2)*(a^2+2ab+b^2)>=64*abc*(a^3+b^3+c^3+3a^2*b+3a^2*c+3b^2*c+3b^2*a+3c^2*a+3c^2*b+6abc) ><==>27cyc(a^4b^2+a^4c^2+2a^3b^3)>=114a^2b^2c^2+10abc*cyc(a^3)+30abc*cyc(a^2b+b^2a) >(***) > >By AMGM the following inequalities hold: >5(a^4b^2+a^4c^2)>=10a^4bc >6(a^4c^2+a^3c^3+b^4a^2+b^4a^2+a^4b^2)>=30a^3b^2c >6(a^4b^2+a^3b^3+c^4a^2+c^4a^2+a^4c^2)>=30a^3c^2b > >Summing the 3 cyclic permutations of each of the above >three, we get >21cyc(a^4b^2+a^2b^4)+12cyc(a^3b^3)>=10abc*cyc(a^3)+30abc*cyc(a^2b+b^2a) >(*) > >Also by AMGM, >6cyc(a^4b^2+a^2b^4)+42cyc(a^3b^3)>=114a^2b^2c^2 (**) > >Adding (*) and (**) gives the desired (***). Let me rewrite it in a more readable way. Denote by cyc(f) the sum cyc(f) = f(a, b, c) + f(b, c, a) + f(c, a, b) 27·(b + c)^{2}·(c + a)^{2}·(a + b)^{2} ³ 64·abc·(a + b + c)^{3} Û 27·(b^{2} + 2bc + c^{2})·(c^{2} + 2ca + a^{2})·(a^{2} + 2ab + b^{2}) ³ ³ 64·abc·(a^{3} + b^{3} + c^{3} + 3a^{2}b + 3a^{2}c + 3b^{2}c + 3b^{2}a + 3c^{2}a + 3c^{2}b + 6abc) Û 27·cyc(a^{4}b^{2} + a^{4}c^{2} + 2a^{3}b^{3}) ³ 114·a^{2}b^{2}c^{2} + 10·abc·cyc(a^{3}) + 30·abc·cyc(a^{2}b + b^{2}a) (***) By AMGM the following inequalities hold: 5·(a^{4}b^{2} + a^{4}c^{2}) ³ 10·a^{4}bc 6·(a^{4}c^{2} + a^{3}c^{3} + b^{4}a^{2} + b^{4}a^{2} + a^{4}b^{2}) ³ 30·a^{3}b^{2}c 6·(a^{4}b^{2} + a^{3}b^{3} + c^{4}a^{2} + c^{4}a^{2} + a^{4}c^{2}) ³ 30·a^{3}c^{2}b Summing the 3 cyclic permutations of each of the above three, we get 21·cyc(a^{4}b^{2} + a^{2}b^{4}) + 12·cyc(a^{3}b^{3}) ³ 10·abc·cyc(a^{3}) + 30·abc·cyc(a^{2}b + b^{2}a) (*) Also by AMGM, 6·cyc(a^{4}b^{2} + a^{2}b^{4}) + 42·cyc(a^{3}b^{3}) ³ 114a^{2}b^{2}c^{2} (**) Adding (*) and (**) gives the desired (***).  No matter what I do, by summing the 3 cyclic permutations of each of the above three, I get the 1st coefficient 5 + 6·4 = 29 and not 21: 29·cyc(a^{4}b^{2} + a^{2}b^{4}) + 12·cyc(a^{3}b^{3}) ³ 10·abc·cyc(a^{3}) + 30·abc·cyc(a^{2}b + b^{2}a) (*) and then (*) and (**) does not add up to (***). What is wrong ??? 

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