Question

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

Answer 1

Do you know the answer? Is it 3/2 ?

Answer 2

yaa

Answer 3

hw did u get it

Answer 4

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

Answer 5

@ganeshie8

Answer 6

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 ?

Answer 7

yes, it is 3/2
http://en.wikipedia.org/wiki/Nesbitt%27s_inequality

Answer 8

but raden i didn't get it

Answer 9

See the link he has posted.

Answer 10

bt i didn't undrstand it

Answer 11

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 :

Answer 12

Now let us consider 3 points on f(x), a,b, and c. Without the loss of generality, let us assume a<b<c.


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.