IIT JEE , AIEEE Forum - Mathematics , Algebra

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

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