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[不等式] 数列最小值问题

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等待hxh posted 2014-4-8 22:55 |Read mode
7.jpg (这一类型的最值问题,有什么套路可循吗?希望高人指点迷津!!)

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其妙 posted 2014-4-8 23:03
回复 1# 等待hxh
kk是全能的,不过不等式却更是他的强项!

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其妙 posted 2014-4-9 00:04
答案:$\dfrac32$,对不对?

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kuing posted 2014-4-9 00:27
回复 3# 其妙

嗯,对的,才想起这个结论:artofproblemsolving.com/Forum/viewtopic.php?p … 020326&#p2020326(5#)

\begin{align*}
&\frac{a_1}{a_2^2+1}+\frac{a_2}{a_3^2+1}+\cdots+\frac{a_n}{a_1^2+1} \\
={}&a_1-\frac{a_1a_2^2}{a_2^2+1}+a_2-\frac{a_2a_3^2}{a_3^2+1}+\cdots+a_n-\frac{a_na_1^2}{a_1^2+1} \\
\geqslant{}& 2-\frac{a_1a_2}2-\frac{a_2a_3}2-\cdots-\frac{a_na_1}2,
\end{align*}
利用贴中的结论有 $a_1a_2+a_2a_3+\cdots+a_na_1\leqslant1$,故原式 $\geqslant3/2$。

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爪机专用 posted 2014-4-9 02:46
这题跟数列根本没关系, 标题没取好。
I am majia of kuing

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original poster 等待hxh posted 2014-4-9 10:08
回复 4# kuing

等号能成立吗? 如果等号成立的话,那就有a1=a2=a3=....an=1,此时与a1+a2+。。。+an=2矛盾

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kuing posted 2014-4-9 12:42
当 $a_1=a_2=1$, $a_3=a_4=\cdots=a_n=0$ 时取等。

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original poster 等待hxh posted 2014-4-9 13:25
哦,对,我以为您用的是均值不等式,所以需要都等于1,看错了,谢谢kuing的精彩解答!

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kuing posted 2014-4-9 13:55
第一步之后是在分母里面用了均值,只看分母中是取不了等的,但由于分子可以为 0,所以还是能取到。

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kuing posted 2014-4-18 17:10
原来我在旧论坛也做过一次
kkkkuingggg.haotui.com/thread-1064-1-5.html
记忆力又差了……

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