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samigupta8 Group Title

if a,b,c>0 the minimum value of a/b+c + b/c+a + c/a+b is

  • one year ago
  • one year ago

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  1. shubhamsrg Group Title
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    Do you know the answer? Is it 3/2 ?

    • one year ago
  2. samigupta8 Group Title
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    yaa

    • one year ago
  3. samigupta8 Group Title
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    hw did u get it

    • one year ago
  4. shubhamsrg Group Title
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    I don;t know the orthodox method for this, but generally,I have learnt that the minimum occurs when a=b=c. I don't know the precise reasoning for it. But I am trying to find out. Anyways, for a=b=c, we get 3/2

    • one year ago
  5. shubhamsrg Group Title
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    @ganeshie8

    • one year ago
  6. shubhamsrg Group Title
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    Here is my attempt : => a/(b+c) + b/(c+a) + c/(a+b) + (b+c)/a + (c+a)/b + (a+b)/c - (b+c)/a - (c+a)/b - (a+b)/c [ a/(b+c) + (b+c)/a ]+ [ b/(c+a) + (c+a)/b ]+ [ c/(a+b) + (a+b)/c ] - [ (b+c)/a + (c+a)/b + (a+b)/c ] For this to be minimum, positive part should be minimum and negative part maximum => 6 - [ (b+c)/a + (c+a)/b + (a+b)/c ] => 6 - [ (b+c)/a +1+ (c+a)/b +1+ (a+b)/c +1 -3] => 6 - (a+b+c)(1/a + 1/b + 1/c) +3 => 9 - (a+b+c)(1/a + 1/b + 1/c) Thus we need to find maximum value of (a+b+c)(1/a + 1/b + 1/c) Now,[ (a+b+c) + (1/a + 1/b + 1/c) ] /2 >= sqrt [ (a+b+c)(1/a + 1/b + 1/c) ] For RHS to be maximum, LHS should be minimum, (a+b+c) + (1/a + 1/b + 1/c) = (a+1/a) + (b+1/b) + (c + 1/c) Minimum vale of this is 6 hence 3>=sqrt [ (a+b+c)(1/a + 1/b + 1/c) ] or 9 >=(a+b+c)(1/a + 1/b + 1/c) My answer then comes out to be 9-9 or 0 hmm, where did I go wrong ?

    • one year ago
  7. RadEn Group Title
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    yes, it is 3/2 http://en.wikipedia.org/wiki/Nesbitt%27s_inequality

    • one year ago
  8. samigupta8 Group Title
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    but raden i didn't get it

    • one year ago
  9. shubhamsrg Group Title
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    See the link he has posted.

    • one year ago
  10. samigupta8 Group Title
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    bt i didn't undrstand it

    • one year ago
  11. shubhamsrg Group Title
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    I was thinking about it and I cam up with a graphical solution. Let us say a+b+c = k , where k is constant for a given a,b and c. We can surely say since a,b,c>0, then k>a ,b and c Now we can re-write our expression as b/(k-b) + a/(k-a) + c/(k-c) Let us consider the graph of x/(k-x) for x<k. f(x) = x/(k-x) f'(x) = k/(k-x)^2 i.e. f'(x) > 0 always, hence f(x) is increasing. f"(x) = 2k/(k-x)^3 i.e. for x<k, f"(x) >0 which means f(x) should be concave upward in the given domain. Graph will be something like this : |dw:1373875648458:dw|

    • one year ago
  12. shubhamsrg Group Title
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    Now let us consider 3 points on f(x), a,b, and c. Without the loss of generality, let us assume a<b<c. |dw:1373875697583:dw| The centroid of this triangle will be (a+b+c /3 ,[ b/(k-b) + a/(k-a) + c/(k-c) ]/3) or since a+b+c = k, centroid is ( k/3 , [ b/(k-b) + a/(k-a) + c/(k-c) ]/3) Consider a vertical line passing through the centroid. It will cut f(x) at (k/3 , (k/3)/(k- (k/3) ) or (k/3 , 1/2) From the figure you can conclude that [ b/(k-b) + a/(k-a) + c/(k-c) ]/3 >= 1/2 or b/(k-b) + a/(k-a) + c/(k-c) >= 3/2 hence its minimum vale is 3/2 and hence our answer. hope I could make this clear.

    • one year ago
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