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Let   \(a,b,c,d\)  be positive real numbers such that \(a+b+c+d=1\) Find the minimum value of

 

\(\frac{a}{b + c + d} + \frac{b}{a + c + d} + \frac{c}{a + b + d} + \frac{d}{a + b + c}.\)
 

 Aug 3, 2019
 #1
avatar+6046 
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using Lagrange multipliers you can show that the minimum occurs when a=b=c=d=1/4 and is equal to 4/3

 

but you can also argue this from symmetry since they are all required to be positive.

 Aug 3, 2019

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