Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How can I prove this inequality

  1. Feb 18, 2012 #1
    I have an inequality and tried to solve it and reached the following:

    Original question: Prove (1/a - 1)(1/b - 1)(1/c - 1) >= 8 when a+b+c = 1 and a,b,c positive

    After expanding and some eliminations,
    I still need to prove 1/a + 1/b + 1/c -1 >= 8

    Any suggestion how to solve it?
  2. jcsd
  3. Feb 18, 2012 #2
    There's no need for anything too fancy, here's how I would tackle it:

    Notice that the conditions on a,b,c are symmetric, that is, consider this a problem of "optimization". Suppose I give you: Maximize x*y*z subject to x+y+z<=100, then you have a few options:

    1 - Solve this problem using LaGrange multipliers
    2 - Realize that this problem is "symmetric" and that whatever x might be, y and z should be the same. In this problem x=y=z=100/3 indeed maximizes the product.

    In the same fashion, what is the largest value that the denominators (a,b,c) will attain? If you can maximize the denominator ( and thus minimize the left-hand side of the inequality), then you'll readily show that the LHS is always greater or equal to 8:



    This will give you a "lower bound for the LHS". It should be pretty easy from here on, does this help?
  4. Feb 18, 2012 #3
    I took the course that had LaGrange multipliers 2 years ago and that course is a prerequisite to this course. So I think your right I should use LaGrange multiplier. Just to check if I am doing to it correct.
    I set f(a,b,c) = 1/a + 1/b + 1/c -1
    and g(a,b,c) = a + b + c -1
    Where 0<a<1,0<b<1,0<c<1

    I solve the 4 equation 4 unknowns:
    fa(a,b,c)=L ga(a,b,c) where fa is derivative of f wrt a...
    fb(a,b,c)=L gb(a,b,c)
    fc(a,b,c)=L gc(a,b,c)

    I get L=-9, a=1/3,b=1/3,c=1/3 (they all interval 0 to 1 thus correct)
    implies f(a,b,c)=8 is either absolute maximum or minimun.
    I take another point and show that its >8 which means 8 is a minimum and solved

    Is this correct solution? Thank you
  5. Feb 18, 2012 #4
    Yep, that's right! If it's a minimum, then picking any other values (Ex: a=0.7, b=0.2, c=0.1) will yield something higher than 9, always. So you're done now. But notice that you could have finished this problem in your head: if a+b+c=1, and the conditions on a,b,c are symmetric (all the same), then one can infer that a=b=c, which implies that 3a=1, or a=b=c=1/3. Does this make sense?

    So a=b=c=1/3 is a maximum for the denominator, or a minimum for the LHS, therefore (1/a)+(1/b)+(1/c) -1 >=8 for all a,b,c>0.

    Good work!
  6. Feb 18, 2012 #5
    Hi there,

    The method with lagrange multipliers can of course be applied, and your solution looks good.

    But the suggestion that local extrema of a symmetric functions are attained when the variables are equal, is just wrong. (It is easy to make counterexamples.)

    To use such argumentation, we can for example rely on Jensens Inequality: When f is convex, then Ʃf(xi) ≥ nf(x), where x is the average of x1,...,xn.

    This can be applied to show 1/a + 1/b + 1/c ≥ 9, with f(x)=1/x.

    Your first inequality can also be shown directly by using Jensen on f(x)=ln(1/x - 1).

    A more elementary solution to your inequality, is to first rearrange to ab+ac+bc≥9abc, then multiply left side by a+b+c, and then rewriting to the obvious true b(a-c)2+a(b-c)2+c(a-b)2≥0.
  7. Feb 18, 2012 #6
    I did not say that "symmetric functions" reach their maxima/minima when all of their values are equal, I said that if the constraints are symmetric, and the function is symmetric with respect to each parameter, then there is no reason that the solution would not be symmetric as well. It's a common argument, but it does indeed break down for asymmetrical constraints/functions.
  8. Feb 18, 2012 #7
    One of the variables could be solved in terms of the other two,

    such as c = 1 - a - b, and that substituted in so that the intended

    inequality to be shown would be in two variables.
  9. Feb 19, 2012 #8
    What you say here is of course correct, but in your posts further up, you seem to apply this to "infer that a=b=c", which does not follow from your above paragraph. It is that inference I claim is not valid, and your reply here does not address that.

    Now, lets make this simple: lets say that you as part of a problem needed to maximize f(x,y)=xy(4x2+4y2-1), with positive x,y, and x+y=10.

    Here everything is symmetric, so f(5,5)=4975 seems to be maximum. Would you accept that solution? If not, how did you infer a=b=c above, when you can not infer x=y here?

    Btw, I find this discussion quite illuminating.
  10. Feb 19, 2012 #9
    Shouldn't I check at x=y and also at the boundaries that is x=0,y=10 and x=10,y=0?
  11. Feb 20, 2012 #10
    Yes, you can check the boundaries (value = 0), and the midpoint, (value = 4975), but my point is that you cannot use symmetry arguments to then conclude that max=4975. Indeed there are two different points at which f attains its maximum value, 4975.03125, and that can be found by lagrange (i suppose), or derivation, or a by rewriting the expression in a clever way.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook