MathLabs

Problem 3

Let x,y,zx,y,z be positive real numbers and let w=xyz3w=\sqrt[3]{xyz}. Prove that (1+x/y)(1+y/z)(1+z/x)≥2+2(x+y+z)/w(1+x/y)(1+y/z)(1+z/x)\ge2+2(x+y+z)/w.
Step 4 of 4: Finish the comparison
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3(x+y+z)/w−1=2(x+y+z)/w+(x+y+z)/w−1≥2(x+y+z)/w+23(x+y+z)/w-1=2(x+y+z)/w+(x+y+z)/w-1\ge2(x+y+z)/w+2
Detailed analysis

Split the lower bound as 3(x+y+z)/w−1=2(x+y+z)/w+(x+y+z)/w−13(x+y+z)/w-1=2(x+y+z)/w+(x+y+z)/w-1. Using (x+y+z)/w≥3(x+y+z)/w\ge3 gives 2(x+y+z)/w+22(x+y+z)/w+2, exactly the required right side.