Let x1, x2, ..., xn be positive reals with sum 1. Show that x1+x2x12+x2+x3x22+⋯+xn+x1xn2≥21.
Solution
Solution:
We have ∑xi+xi+1xi2−∑xi+xi+1xi+12=∑(xi−xi+1)=0. Also, xi+xi+1xi2+xi+12≥21(xi+xi+1) by the QM-AM inequality. Summing over all i, ∑xi+xi+1xi2+∑xi+xi+1xi+12≥21∑(xi+xi+1)=1. Since the two sums are equal, each is at least 21. Therefore, ∑xi+xi+1xi2≥21.
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