Let x1,x2,…,xn be positive numbers such that: x1n−1+x2n−1+⋯+xnn−1=x1x2…xn Prove the inequality: (x1−n+1)(x2−n+1)…(xn−n+1)≥1.(Serdiuk Nazar)
Solution
Using Cauchy inequality we get ∀i=1,n: x1x2…xn=x1n−1+x2n−1+⋯+xnn−1≥x1n−1+(n−1)x1x2…xn−1xn+1…xn⇒x1x2…xn−1xn+1…xn(xi−n+1)≥xin−1⇒xi−n+1≥x1x2…xn−1xn+1…xnxin−1⇒i=1∏n(xi−n+1)≥i=1∏nx1x2…xn−1xn+1…xnxin−1=1, as needed.
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