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A new problem

  1. Feb 1, 2007 #1
    the problem statement is:
    if a,b,c are real numbers such that [tex]\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = 2[/tex]

    we have to prove that:
    [tex]\frac{1}{4a+1} + \frac{1}{4b+1} + \frac{1}{4c+1} \geq 1[/tex]

    thanks in advance.
  2. jcsd
  3. Feb 1, 2007 #2

    Gib Z

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    Well I thought id at least prove a general class of the solutions, where a=b=c, but it turns out that the only solution is a=b=c=0.5. It is the only solution where the 2nd equtation is exactly 1. The others must be more.

    I tried another case, where 2 solutions (a and b) are equal, and it shows a=1/(2c+1). That might help, perhaps we could prove every case to be true.
  4. Feb 1, 2007 #3

    Hence,the condition is equivalent with:

    By similar simple algebra the inequality can be rewritten as follows:
    [tex]\frac{4a}{4a+1}+\frac{4b}{4b+1}+\frac{4c}{4c+1}\leq 2 [/tex]

    Now,let's consider the difference "Inequality-Condition":

    [tex]\left(\frac{4a}{4a+1}-\frac{a}{a+1}\right)+\left(\frac{4b}{4b+1}-\frac{b}{b+1}\right)+\left(\frac{4c}{4c+1}-\frac{b}{b+1}\right)\leq 1[/tex]

    Every expression within brackets is of form [itex]\frac{4x}{4x+1}-\frac{x}{x+1}[/itex].

    [tex]\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}?\rightarrow 4x^2-4x+1<0\rightarrow 4\left(x-\frac{1}{2}\right)^2<0\longrightarrow Contradiction![/tex]


    EDIT:BTW,this cute little problem was misplaced in dumb brainteasers section.
    Elementar proof presented here is understandable even to 6th graders.:smile:
    Last edited: Feb 1, 2007
  5. Feb 3, 2007 #4


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    That's simply a brilliant solution. Wow !:approve: :cool:
  6. Feb 4, 2007 #5
    how do you get a contradiction here? i get:

    [tex]\frac{4x}{4x+1} - \frac{x}{x+1} > \frac{1}{3}[/tex]

    [tex]\frac{3x}{(4x+1)(x+1)} > \frac{1}{3}[/tex]

    [tex]\frac{9x}{(4x+1)(x+1)} > 1[/tex]

    [tex]\frac{9x}{(4x+1)(x+1)} - 1 > 0[/tex]

    [tex]\frac{9x - 4x^2 - 5x - 1}{(4x+1)(x+1)} > 0[/tex]

    [tex]\frac{4x^2 - 4x + 1}{(4x+1)(x+1)} < 0[/tex]

    [tex]\frac{(2x-1)^2}{(4x+1)(x+1)} < 0[/tex]

    [tex](4x+1)(x+1) < 0[/tex]

    [tex]x \in \left(-\frac{1}{4}, -1\right)[/tex]

    so where is the contradiction?
    Last edited: Feb 4, 2007
  7. Feb 4, 2007 #6

    Gib Z

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    Ahh heres your problem murshid_islam, you know when you copyed this problem over from the brain teasers section? https://www.physicsforums.com/showthread.php?t=131366

    Take a look at the original question, a b and c have to be POSTIVE. All the solutions to that quadratic inequality are negative, and therefore we reach a contradiction.
  8. Feb 4, 2007 #7
    thanks Gib Z. but still, the contradiction tehno got is wrong, isn't it?
  9. Feb 4, 2007 #8

    Gib Z

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    Why is the contradiction wrong? The inequaility has only negative solutions, which is contradictory from the original assumption of positive solutions.
  10. Feb 4, 2007 #9
    how do you get from [tex]\frac{4x}{4x+1}-\frac{x}{x+1}>\frac{1}{3}[/tex] to [tex]4x^2-4x+1<0[/tex]

    thats what i said was wrong?
  11. Feb 4, 2007 #10

    Gib Z

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    Well ive tried for 30 mins now, i cant see how he got that either. Help tehno!
  12. Feb 4, 2007 #11


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    From here, x is positive, and so the denominator is always positive. So, for the inequality to hold, we require [itex]4x^2 - 4x + 1<0[/itex]. From here, the contradiction is obtained as in tehno's post.
    Last edited: Feb 4, 2007
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