if a,b,c>0 the minimum value of a/b+c + b/c+a
+ c/a+b is
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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
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 ?
yes, it is 3/2
but raden i didn't get it
See the link he has posted.
bt i didn't undrstand it
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 0 always, hence f(x) is increasing.
f"(x) = 2k/(k-x)^3
i.e. for x0 which means f(x) should be concave upward in the given domain.
Graph will be something like this :
Now let us consider 3 points on f(x), a,b, and c. Without the loss of generality, let us assume a= 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.