MathLabs

Problem 5

Determine all possible values of S=aa+b+d+ba+b+c+cb+c+d+da+c+dS=\dfrac{a}{a+b+d}+\dfrac{b}{a+b+c}+\dfrac{c}{b+c+d}+\dfrac{d}{a+c+d}, where a,b,c,da,b,c,d are arbitrary positive real numbers.
Step 1 of 4: Prove the lower bound
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S>a+b+c+da+b+c+d=1S>\dfrac{a+b+c+d}{a+b+c+d}=1
Detailed analysis

Replacing every denominator by the larger sum a+b+c+d decreases each positive fraction. The resulting four fractions add to 1, so S is strictly greater than 1.