MathLabs

Problem 2

Let a,b,c be positive real numbers with abc=8. Prove that a squared divided by the square root of (1+a cubed)(1+b cubed), plus the two cyclic analogues, is at least 4/3.
Step 3 of 5: Combine the rational terms
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S=2(a2+b2+c2)+(ab)2+(bc)2+(ca)2,D=72+2SS=2(a^2+b^2+c^2)+(ab)^2+(bc)^2+(ca)^2,\quad D=72+2S
Detailed analysis

Putting the three rational terms over the common denominator D=(2+a squared)(2+b squared)(2+c squared), and using a squared b squared c squared=64, their numerator is 2S and D=72+2S. Hence the lower bound is S/(36+S).