已知xi∈R+,i=1,2,…,n 求证不等式n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)已知xi∈R+,i=1,2,…,n求证不等式 n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)等号当且仅当x1=x2=…=xn时取得这个题是好像是可以通

已知xi∈R+,i=1,2,…,n 求证不等式n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)已知xi∈R+,i=1,2,…,n求证不等式 n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)等号当且仅当x1=x2=…=xn时取得这个题是好像是可以通
已知xi∈R+,i=1,2,…,n 求证不等式n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)
已知xi∈R+,i=1,2,…,n
求证不等式 n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)
等号当且仅当x1=x2=…=xn时取得
这个题是好像是可以通过适当的构造，然后可以用均值不等式证明的
希望各位大侠给的证明可以清晰明快啦……

已知xi∈R+,i=1,2,…,n 求证不等式n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)已知xi∈R+,i=1,2,…,n求证不等式 n/(n+1)≥x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)等号当且仅当x1=x2=…=xn时取得这个题是好像是可以通
我看不懂2楼的答案,下面是我的解答(看图片）.并附上latex代码.
设$a_1=\frac{x_2}{x_1},a_2=\frac{x_3}{x_2},\cdots,a_2=\frac{x_1}{x_n}$,则我们有$a_1a_2\cdots a_n=1$,我们要证明的是:$\frac{1}{n+a_1}+\frac{1}{n+a_2}+\cdots+\frac{1}{n+a_n}\le \frac{n}{n+1}$  (1)
而由Cauchy不等式得到:
\[(n+a_1)(\frac{1}{n-1+a_1}+\frac{1}{n^2}) = (n-1+a_1+1)(\frac{1}{n-1+a_1}+\frac{1}{n^2}) \ge \left(\frac{n+1}{n}\right)^2\]
所以我们有：
\[\frac{1}{n+a_1} \le \left(\frac{n}{n+1}\right)^2 (\frac{1}{n-1+a_1}+\frac{1}{n^2}) \]
同理我们有,
\[\frac{1}{n+a_2} \le \left(\frac{n}{n+1}\right)^2 (\frac{1}{n-1+a_2}+\frac{1}{n^2}) \]
$\cdots$
\[\frac{1}{n+a_n} \le \left(\frac{n}{n+1}\right)^2 (\frac{1}{n-1+a_n}+\frac{1}{n^2}) \]
因此为证明不等式(1)成立我们只需证明:$\frac{1}{n-1+a_1}+\frac{1}{n-1+a_2}+\cdots+\frac{1}{n-1+a_n}\le 1$(2)
而(2)等价于$\frac{a_1}{n-1+a_1}+\frac{a_2}{n-1+a_2}+\cdots+\frac{a_n}{n-1+a_n}\ge 1$(3)
而(3)式可由局部不等式
$\frac{a_1}{n-1+a_1} \ge \frac{a_1^k}{a_1^k+a_2^k+\cdots+a_n^k}$(4)
其中$k=\frac{n-1}{n}$,(4)式的证明是由于$(4) \iff a_1(a_2^k+\cdots+a_n^k) \ge (n-1)a_1^k$而最后一式有平均值不等式可得到.
同理
$\frac{a_1}{n-1+a_1} \ge \frac{a_2^k}{a_1^k+a_2^k+\cdots+a_n^k}$
$\cdots$
$\frac{a_n}{n-1+a_1} \ge \frac{a_n^k}{a_1^k+a_2^k+\cdots+a_n^k}$
把以上各式相加便得到了(3)从而证明了原命题.

-a: The first set of each root zone, the second document location (70% of the root line with this principle, so look first know in which area among the five keys to see The second document on the posi...

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-a: The first set of each root zone, the second document location (70% of the root line with this principle, so look first know in which area among the five keys to see The second document on the positioning in the 5 key in which! do not believe? root distribution can be above all the root try again by this principle,cheap yves saint laurent shoes, the special ring out to see, not much right, only the following 25% +5%, and then removed with a special memory methods,manolo blahnik shoes sale, only a few, just remember it!) -

2: \The vertical position (No. 2 key, F key) -

- not the back I decided, as long as we removed several times, soon learned the legendary difficult to learn the five strokes of the! And do not start learning English key! -
-
- now part of streamlining the learning into: 5 minutes theory -> practice Separation of Chinese -> removed the root -> the first set area -> second tranche of positioning -> OK! (No need rote lifetime, fast and accurate!) -
-
2: to know the five strokes of the keyboard on the distribution of characteristics: recognized root of the first and second stroke on the set location -
formulas added back this difficult period, it is the most failure of an inventor! The abstract image of what was so complicated and difficult to remember! Alas! Unfortunately,) -
original ideas sum up:
Just remember the following 3 points: -
I must learn to play five strokes back pain: theory -> back I decided -> Training Separation of Chinese -> removed the root -> back I decided -> NOT true I must -> corresponds to Which English button -> OK! (Interrupted at any time due to pain) -
- (invented the five-stroke, great! really is re-code at least,christian louboutin boots, the fastest input method! -
1: \The horizontal position (No. 1 key, G key) -
b: the root of a special memory method (25% of the root): If \
c: can only memorize the root (only 5%): such as \
-
3: \write bit (No. 3 keys, D keys, but also pronounced \
...... -
-
found that 70% of the root principles do not need to remember it, and the rest is very easy to think up ways to remember! -
-
good! actual look: -
- they simply do not have back pain that I must, skip it and go directly to the Separation of Chinese stage right,christian louboutin flats uk, if a mouth mutter prayers, may wish to \for the words must come with it! -
-
1: know what Wubi is: 1 丨 Pie Dian B (if they had written strokes off) -
(red strokes shown in Figure District: Bank on the left 5 keys of \
up on the left 5 keys of \
down the left five health is \
(blue strokes shown in Figure bit: to the middle yellow line as the boundary, from the middle to both sides of the walk, are \
3: Distribution of root principles: (inventors by the idea that to accommodate the root of so much of) -
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Her husband's confession

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这类的题应该是数学竞赛的吧

可采用双螺旋数学归纳法来证 不过比较难想

或者用琴声不等式

具体如下 

楼主 只要稍加变换即可

证明：
首先利用分析法容易证明，对任意正实数x,y有
[1/(n+x)+1/(n+y)]/2<=1/(n+(xy)/2)
然后利用琴生不等式即可推广到n个变量的情况。
x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)
=1/(n+x2/x1)+1/(n+x3/x2)+…+1/(n+x1/xn)
<=n/(n+(x2/x1)(...

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证明：
首先利用分析法容易证明，对任意正实数x,y有
[1/(n+x)+1/(n+y)]/2<=1/(n+(xy)/2)
然后利用琴生不等式即可推广到n个变量的情况。
x1/(nx1+x2)+x2/(nx2+x3)+…+xn/(nxn+x1)
=1/(n+x2/x1)+1/(n+x3/x2)+…+1/(n+x1/xn)
<=n/(n+(x2/x1)(x3/x2)……(x1/xn))
=n/(n+1)
等号成立当且仅当x1=x2=……=xn

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