(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 <= x_1, x_2, ..., x_n <= b where both a and b are positive and a < b: (x_1 + x_2 + ... + x_n) (1/x_1 + 1/x_2 + ... + 1/x_n) <= n^2 (a+b)^2 /(4ab)

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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 <=1+1/b due to (x-1)(b-x)>=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/b+(1+b)(n-2)(1+b)/b+(n-2)^2(1+b)^2/(4b)
=(1+b)^2/(4b)(4n-4+n^2-4n+4)
=(n^2(1+b)^2/(4b)