Given x, y, and z are real numbers that satisfy x+y+z=x1+y1+z1 and xyz=1. Find the smallest value of ∣x+y+z∣.
My Way
Based on QM≥AM,
x1+y1+z13≥3x+y+z9≥(x1+y1+z1)(x+y+z)
Because x+y+z=x1+y1+z1,
9≥(x1+y1+z1)(x+y+z)9≥(x+y+z)2±3≥x+y+z3≥∣x+y+z∣
But, I found the maximum value. Can you give me a hint?
#Algebra
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Seems like 1 is the minimum value. However, I'm stuck atm.