Let a1,a2,…,a12 be real numbers such that a1+a2+⋯+a12=1. Find the minimum value of a12+a22+⋯+a122.
Solution — click to reveal
By Cauchy-Schwarz, (12+12+⋯+12)(a12+a22+⋯+a122)≥(a1+a2+⋯+a12)2,so a12+a22+⋯+a122≥121.Equality occurs when ai=121 for all i, so the minimum value is 121.