Prove for all a,b,c>0: a/(b+c) + b/(a+c) + c/(a+b) >= 3/2 ?

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The inequality a/(b+c) + b/(a+c) + c/(a+b) ≥ 3/2 holds true for all positive real numbers a, b, and c. A suggested approach to prove this involves substituting x = b+c, y = a+c, and z = a+b, allowing the left side to be rewritten as 1/2 [(x/y + y/x) + (y/z + z/y) + (z/x + x/z) - 3]. This transformation simplifies the proof process and highlights the symmetry in the variables.

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barbiemathgurl
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can somebody prove that for all a,b,c>0:

a/(b+c) + b/(a+c) + c/(a+b) >= 3/2
 
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There are probably a few ways you can do this.

Here's a hint: Set x=b+c, y=a+c, and z=a+b. Then re-write the left side as: 1/2 [(x/y + y/x) + (y/z + z/y) + (z/x + x/z) - 3].
 

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