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Prove:if 1/a+1/b+1/c=1 and a,b,c >0 then (a-1)(b-1)(c-1)>=8

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  1. Oct 8, 2015 #1
    The problem: prove that if 1/a + 1/b + 1/c=1 and a,b,c are positive numbers then (a-1)(b-1)(c-1)>=8

    I've tried it myself but couldn't do it. I have tried going backwards i.e. :
    (a-1)(b-1)(c-1)>=8
    (fast forward) abc -(ab+bc+ac) +a+b+c -1 >=8
    Now from 1/a + 1/b + 1/c = 1 we have that ab+bc+ac/abc=1 thus :
    ab + bc + ac = abc , so abc -(ab+bc+ac) +a+b+c -1>=8 is really a+b+c -1>=8
    that is a+b+c >=9 . So If I could somehow prove that a+b+c>9 then I could work backwards and prove the former statement.
    But the only progress I've made is :all of a,b,c must be greater than 1 because if they weren't then any of the 1/a,1/b,1/c would be greater then 1 thus we would need negatives(and a,b,c are positive) to "bring back" the value to 1, nor can they equal to 1 because the same thing would happen(one of 1/a,1/b,1/c would be equal to 1 and we would need negatives)so a>1, b>1, c>1 adding these three together we get a+b+c >3 ,and I need a+b+c+>=9.
    Sorry for the long post. Could anyone give me some directions?

    Thanks,
    Daniel
     
  2. jcsd
  3. Oct 8, 2015 #2

    RUber

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    Could you prove it for 2?
    ## \frac1a + \frac1b = 1## Prove that ##ab\geq 4##?
     
  4. Oct 8, 2015 #3

    RUber

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    Your problem hinges on the relationship between a, b, and c.
    If a = b = c, then they are all 3, right?
    Let's change one of the three...let a be (3-x), either b or c would have to increase to maintain the 1/a + 1/b + 1/c = 1 balance.
    How much would it change?
    Let's say they absorb the difference equally...share the burden, then you have ##\frac{1}{3-x} + \frac{2}{3+y} = 1##. Can you solve for y in terms of x? If y > x/2 you have shown that (3-x) + (3+y)+(3+y) > 9.
     
  5. Oct 8, 2015 #4
    Thank you RUber!!!
     
  6. Oct 8, 2015 #5

    Ray Vickson

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    You can regard this as a constrained optimization problem:
    [tex] \begin{array}{l} \min_{a,b,c} f(a,b,c) = (a-1)(b-1)(c-1) \\
    \text{subject to} \\
    \displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1
    \end{array}
    [/tex]
    Let us restrict the problem to ##a,b,c > 1##. There is a good reason for this, because if you allow any of ##a,b,c## to be ##< 1## but positive, then you must also allow one or more of them to be ##< 0## in order to balance out the constraint. In that case you can construct a sequence of feasible ##a_n, b_n, c_n## values that give ##f(a_n,b_n,c_n) \to -\infty##: an unbounded problem having no minimum.

    OK, so we want ##a, b, c > 1##. Solve the constraint for ##c## as a function of ##a,b##, then plug that formula into ##f(a,b,c)## to get a new function ##F(a,b)## involving ##a## and ##b## alone, and with no constraint anymore (other than ##a,b > 1##, and implicitly, ##c > 1##). This unconstrained 2-variable optimization problem can be solved by finding the stationary point (or points) of ##F(a,b)##. If ##\min_{a,b} F(a,b) \geq 8## you are done.
     
  7. Oct 9, 2015 #6
    Wow, I would've never thought of something like that. Although this way of proving is a little too complex for me.
    Thank you for your reply!
     
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