2
|
9433 Views
|
2 Replies
|
9 Total Likes
View groups...
Share
GROUPS:

A Hands-on Approach to Wei's Inequality

Posted 12 years ago
 The idea of this thread is from a question raised by Dongyi Wei, who is a very competitive first tier math olympiad in China. He proposed the following inequality should hold for positive real numbers a, b and c: He sent this problem to his buddy to seek a solution in terms of elementary and algebraic method, which is very lengthy and tricky. Since I am more engineer than a math athlete, I do not think I can come up with that tricky solution and I just want things "work". To make the problem easier to describe, I defined two functions for each side of the inequality: f is for the lhs f[a_, b_, c_] = (1 - a)^2 + (1 - b)^2 + (1 - c)^2;and g is for the rhsg[a_, b_,  c_] = (1 - a^2) (1 - b^2) c^2/(a b + c)^2 + (1 - b^2) (1 - c^2) a^2/(b c + a)^2 + (1 - c^2) (1 - a^2) b^2/(c a + b)^2;(It is preferred to use Set instead of SetDelay(:=) in this case)My first attempt was a simple brutal-force method, I just used Maximize[{g[a, b, c]/f[a, b, c], a > 0, b > 0, c > 0}, {a, b, c}]which seemed to take  forever to return the expected result, 1. This definitely does not work. Also I realize that I cannot evern visualize f and g because they have three variables and their plot would require the 4th dimension.As I read the problem twice, I noticed that if this inequality holds and I gave f[a,b,c] a fixed value, then this value should be the max value of the rhs. Aha! That sounds like I can study this problem by the contours of f and g. Based on the form of the two function f and g, it is easier to find the proper a, b and c for f than for g if a fixed value is given, Actually, f[a, b,c ] == constant yields a standard form for a 2-D sphere. I can quickly parameterize a, b and c here using spherical coordinates: (upper half is removed for demonstration, r is the radius of each sphere)ParametricPlot3D[ Table[{r*Cos[?] Cos[?] + 1, r*Cos[?] Sin[?] + 1, r*Sin[?] + 1}, {r, 1, 4}], {?, 0, 2 ?}, {?, -?, 0}, AxesLabel -> {a, b, c}, LabelStyle -> {Italic, FontSize -> 15}, Mesh -> None, Boxed -> False] While g is super complicated: Show[Sequence @@ (ContourPlot3D[      g[a, b, c] == #, {a, 0.1, 4}, {b, 0.1, 4}, {c, 0.1, 4},       AxesLabel -> {a, b, c}, LabelStyle -> {Italic, FontSize -> 15},       Contours -> 3, Mesh -> None, Boxed -> False] & /@ {0.1, 1, 2, 3})]Note that this is still a plot for the contour of f and g in terms of a, b and c. If I plot f against theta and phi, it is just a plane with f = r^2 in the theta-phi-f coordinate. After I set a value for f, it is also required that the values of a, b and c in the rhs must be the same set of a, b and c used in the lhs. Thus I can just use the parameterized a, b and c in the function g, ie. convertingg[a,b,c]tog[r*Cos[\[Phi]] Cos[\[Theta]] + 1, r*Cos[\[Phi]] Sin[\[Theta]] + 1, r*Sin[\[Phi]] + 1]Now I can plot the g against theta and phi with Plot3D as r is set. So what I am looking for now is whether the maximum value of g exceeds r^2, a preset value for f. For some r's (from 0.5 to 5,  the step length be 0.5), I can find related surfaces (a quater is removed for demonstration purpose): Plot3D[Table[ With[{r = r},              g[r*Cos[?] Cos[?] + 1, r*Cos[?] Sin[?] + 1,r*Sin[?] + 1]], {r, 0.5, 5, 0.5}             ],          {?, 0, ?/2}, {?, 0, ?/2}, AxesLabel -> {?, ?, "g"}, LabelStyle -> {Italic, FontSize -> 15},          PlotRange -> Full, Mesh -> None,          RegionFunction -> Function[{x, y, z}, ?/4 < Abs[x - ?/2] + Abs[y - ?/4]], Boxed -> False ]Just for convience, I can define a functional h: h[r_] = g[r*Cos[\[Phi]] Cos[\[Theta]] + 1, r*Cos[\[Phi]] Sin[\[Theta]] + 1, r*Sin[\[Phi]] + 1]So the maximum value of each slide of this square dumpling is NMaximize[h,...]. For instance, when r  is 2, I haveIn[5]:= NMaximize[{Evaluate[h[2]], 0 < \[Theta] < \[Pi]/2, 0 < \[Phi] < \[Pi]/2}, {\[Theta], \[Phi]}]Out[5]= {4., {\[Theta] -> 0.785398, \[Phi] -> 0.61548}}I can also tabulate the results for several r's: In[6]:= Table[            {i, NMaximize[{Evaluate[h[i]], 0 < \[Theta] < \[Pi]/2, 0 < \[Phi] < \[Pi]/2}, {\[Theta], \[Phi]}][[1]]},             {i, 0.5, 5, 0.2}       ] // TableForm[#, TableHeadings -> {None, {"r", "g_Max"}}] &We can see that the maximum value of g is clearly r^2 and Wen's proposal at least is emperically correct.
2 Replies
Sort By:
Posted 12 years ago
 I think it is hard to do this algebraically.  I like Shegnui's bold approach of trying to draw a contour plot.Another empirical approach is to take on the inequality for particular values of c.  Then the algebraic approach is manageable, and so is the calculus.For example, for c=1, there is one local minimum in the domain of positive values, {a,b}={1,1}  and trying other values of c that I tried the only local minimum is at {c,c}, e.g. here with c=2Note the inequality is an equality at a=b=c, and then the boundary case when one of the variables is -, e.g. c=0, is a trivial inequality.  But I wasn't able to get it it to return for the symbolic c, so this is just empirical evidence.
Posted 12 years ago
 Why didn`t you try to bring everything to the left hand side and to simplify it out? You need an expression manifestly bigger than zero then. A square of something or a sum or product of squares. Give it a try.