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 5 of 5: Finish the inequality
S36+S≥7236+72=43\dfrac{S}{36+S}\ge\dfrac{72}{36+72}=\dfrac43
Detailed analysis

The function S over (36+S) is increasing for positive S. Since S is at least 72, the lower bound is at least 4/3. Equality occurs at a=b=c=2.