IIT JEE , AIEEE Forum - Mathematics , Algebra

elastiboysai

Lemme see who solves it first

Determine range of

[ a/(a+b+d) + b/(a+b+c) + c/(b+c+d) + d/(a+c+d) ] for positive reals a, b, c, d.
This is very simple

computer001

<=4/3 ?

hash_include

i think the min tends to 1 and max tends to 2??

i.e range is (1,2)

computer001

pl temme how it tend to 2

hash_include

let a>b>c>d
b+c+d < b + c+ a
$=> \frac b {b+c+a} + \frac d {a+c+d} <= \frac b {b+c+d} + \frac d {b+c+d}$
add $\frac c {b+c+d}$ on both sides and the RHS will be 1
now, $0< \frac a {a+b+d} < 1$
adding that on both sides, we get the inequality Smile

NOTE: i did this same sum very recently, which is y i told d ans immediately Mr. Green

sboosy

computer001

edited

but i think there shud be a stricter inequality

hsbhatt

I am not sure whether you are asking the range in terms of a,b,c and d or actual values.

greater than 1 is easily proved.

Also,

$\frac{a} {a+b+d} + \frac{b} {a+b+c} + \frac {c} {b+c+d} + \frac {d} {a+c+d} \leq \frac {a} {a+b} + \frac {b} {a+b} + \frac {c} {c+d} + \frac {d} {c+d} = 2$

Sorry, that should be a strict inequality.

But if it is in terms of a,b,c and d, we must note that if a>b>c>d, then

$\frac {1} {b+c+d} \geq \frac {1} {a+c+d} \geq \frac {1} {a+b+d} \geq \frac {1} {a+b+c}$

So, if we call the given expression S, Rearrangement Inequality gives,

$S \leq \frac {a} {b+c+d} + \frac {b} {a+c+d} + \frac {c} {a+b+d} + \frac {d} {a+b+c}$

and $S \geq \frac {d} {b+c+d} + \frac {c} {a+c+d} + \frac {b} {a+b+d} + \frac {a} {a+b+c}$

hsbhatt

also, because of the homogoneity of the expression we can assume a+b+c+d = 1.

I think smoothing principle would give you the minimum as 4/3. Could someone pls confirm this?

pardesi

no that's not true S can be considered as a function in each variable seperately holding out others constant and in each case it is continiouas at each point in the reals - the point where it blows up
also it's clear that 1<=S<=2
we now prove that one can get as close one wants to both of them
set a=b=u and c=d=v
now take the limit x--->1 and y--->0 along any path u want(if u wnat them to go depenedently otherwise no prob)(as the cauchy-reimann conditions hold true)
we get the limit as 1
set a=c=u and b=d=v and take limits as u--->1 and v--->0if u wnat them to go depenedently otherwise no prob) we get limit as 2
so we can get arbitrarily close

elastiboysai

Ya thats rite
Ans is 1<=A<=2

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