Let n be a positive integer and let x1,x2,…,xn be positive real numbers such that x1x2⋯xn=1. Prove that i=1∑nxin(1+xi)≥2n−1ni=1∏n(1+xi).
Solution
By the power-mean inequality, 1+an≥2n−1(1+a)n,a≥0.(∗) Thus, i=1∑nxin(1+xi)=i=1∑nxin+i=1∑nxin+1≥i=1∑nxin+n(i=1∏nxi)1+1/n(AM-GM)=i=1∑nxin+n=i=1∑n(1+xin)≥2n−11i=1∑n(1+xi)nby (*)≥2n−1ni=1∏n(1+xi).(AM-GM) Clearly, equality holds if and only if x1=x2=⋯=xn=1.
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Source: MathNet,
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