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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
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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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
Do you know the answer? Is it 3/2 ?
 one year ago

samigupta8 Group TitleBest ResponseYou've already chosen the best response.0
hw did u get it
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
@ganeshie8
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
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 99 or 0 hmm, where did I go wrong ?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
yes, it is 3/2 http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
 one year ago

samigupta8 Group TitleBest ResponseYou've already chosen the best response.0
but raden i didn't get it
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
See the link he has posted.
 one year ago

samigupta8 Group TitleBest ResponseYou've already chosen the best response.0
bt i didn't undrstand it
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
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 rewrite our expression as b/(kb) + a/(ka) + c/(kc) Let us consider the graph of x/(kx) for x<k. f(x) = x/(kx) f'(x) = k/(kx)^2 i.e. f'(x) > 0 always, hence f(x) is increasing. f"(x) = 2k/(kx)^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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
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/(kb) + a/(ka) + c/(kc) ]/3) or since a+b+c = k, centroid is ( k/3 , [ b/(kb) + a/(ka) + c/(kc) ]/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/(kb) + a/(ka) + c/(kc) ]/3 >= 1/2 or b/(kb) + a/(ka) + c/(kc) >= 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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