MathLabs

Problem 2

Let a,b,ca,b,c be positive real numbers with abc=1abc=1. Prove that 1a3(b+c)+1b3(c+a)+1c3(a+b)≥32\frac1{a^3(b+c)}+\frac1{b^3(c+a)}+\frac1{c^3(a+b)}\ge\frac32.
Step 3 of 3: Finish with AM–GM
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In plain words

The product constraint supplies the final numerical lower bound.

x+y+z≥3(xyz)1/3=3x+y+z\ge3(xyz)^{1/3}=3
Detailed analysis

By AM–GM, x+y+z≥3(xyz)1/3=3x+y+z\ge3(xyz)^{1/3}=3. Combining this with E≥x+y+z2E\ge\frac{x+y+z}{2} yields E≥32E\ge\frac32, exactly the required inequality.