(x_1+...+x_n) (1/x_1+...+1/x_n)<=n^2(a+b)^2/(4ab)

 来源: haha2000 于 06-12-08 11:42:26 Prove the inequality for numbers a
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Deuss: induction works, but a little clumsy

can assume a=1, otherwise dividing all xi by a.
Let xn be the max one, then change xn to b will not decrease xn/x+x/xn.
Let x[n-1] be the min one, then change x[n-1] to 1 will not decrease x[n-1]/x+x/x[n-1].
Plus, 1/x+x/b =0.
All these combined plus induction hypothesis, we get
the product (SUM_n x[i])(SUM_n 1/x[i])(1+b)(1+1/b)+(1+b)(SUM_{n-2} 1/x[i]) + (1+1/b)(SUM_{n-2} x[i])+(SUM_{n-2} 1/x[i])(SUM_{n-2} x[i])
=(1+b)^2/(4b)(4n-4+n^2-4n+4)
=(n^2(1+b)^2/(4b)