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 4 of 5: Bound S by AM-GM
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S≥2⋅3(abc)2/3+3((ab)2(bc)2(ca)2)1/3=24+48=72S\ge2\cdot3(abc)^{2/3}+3((ab)^2(bc)^2(ca)^2)^{1/3}=24+48=72
Detailed analysis

AM-GM gives a squared+b squared+c squared at least 12 because abc=8, and (ab) squared+(bc) squared+(ca) squared at least 48. Therefore S is at least 72, with equality only when a=b=c=2.