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文[1]中证明了如下不等式: 若α_i>0(i=1,2,…,n),则multiply from i=1 to n(1+a_i)≥(1+multiply from i=1 to n a_i~)(1/n))~n;(1) 若x_i>0,sum from i=1 to n x_i=1,则 multiply from i=1 to n(x_i+1/x_i)≥(n+1/n)~n(2) 本文拟利用函数的单调性给出(1)与(2)的推广及简便证法。引理1 设p,q,α,β∈R~+,且pα≥≥qβ,则f(x)=αx~p+βx~(-g)。是[1,+∞)内的单调递增函数。证当x∈[1,+∞)时, f′(x)=x~(p-1)[pα-qβx~(-(p+q))]≥0故引理得证。引理2 设αi,p_i∈R~+,(i=1,2,…,n),∑p_i=1,则 f(x)==∏(x+α_i)~(pi)-x是[0,+∞)内的单调递增函数,当且仅当α_1=α_2=…=α_n时,f(x)不是严格递增的,这时f(x)为常数。
The following inequality is proved in [1]: If α_i>0(i=1,2,...,n), multiply from i=1 to n(1+a_i)≥(1+multiply from i=1 to n A_i~)(1/n))~n;(1) If x_i>0,sum from i=1 to n x_i=1, then multiply from i=1 to n(x_i+1/x_i)≥(n+ 1/n)~n(2) In this paper, we will use the monotonicity of functions to give the generalizations and simplifications of (1) and (2). Lemma 1 Let p, q, α, β∈R~+, and pα≥≥qβ, then f(x)=αx~p+βx~(-g). Is a monotonically increasing function within [1,+∞). When x∈[1,+∞) proves that f′(x)=x~(p-1)[pα-qβx(-(p+q))]≥0, the Lemma gains. Lemma 2 Let αi,p_i∈R~+,(i=1,2,...,n),∑p_i=1, then f(x)==∏(x+α_i)~(pi)-x is[ The monotonically increasing function within 0,+∞), if and only if α_1=α_2=...=α_n, f(x) is not strictly increasing, in which case f(x) is a constant.