Cho \(a_1\le a_2\le....\le a_n\) thỏa mãn \(\hept{\begin{cases}a_1+a_2+a_3+...+a_n=0\\\left|a_1\right|+\left|a_2\right|+\left|a_3\right|+...+\left|a_n\right|=1\end{cases}}\)
CMR: \(a_n-a_1\ge\frac{2}{n}\)
Cho \(\hept{\begin{cases}a_1>a_2>...>a_n>0\\1\le k\in Z\end{cases}}\)
CMR : \(a_1+\frac{1}{a_n\left(a_1-a_2\right)^k\left(a_2-a_3\right)^k...\left(a_{n-1}-a_n\right)^k}\ge\frac{\left(n-1\right)k+2}{\sqrt[\left(n-1\right)k+2]{k^{\left(n-1\right)k}}}\)
với \(a_1,a_2,a_3,.....,a_n>0;a_1+a_2+a_3+....+a_n=k\)
Chứng minh\(\left(a_1+\frac{1}{a_2}\right)^2+\left(a_2+\frac{1}{a_3}\right)^2+...+\left(a_n+\frac{1}{a_1}\right)^2\ge\frac{1}{n}\left(\frac{k^2+n^2}{k}\right)^2\)
(Nghi binh 20/09)
Cho \(a_1,a_2,...,a_n>0;3\le n\in N.\) Đặt:
\(A_1=\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+...+\frac{a_{n-1}}{a_n+a_1}+\frac{a_n}{a_1+a_2}\)
\(A_2=\frac{a_1}{a_n+a_2}+\frac{a_2}{a_1+a_3}+...+\frac{a_{n-1}}{a_{n-2}+a_n}+\frac{a_n}{a_{n-1}+a_1}\)
Chứng minh rằng: \(Max\left\{A_1,A_2\right\}\ge\frac{n}{2}\)
Chứng minh rằng với các số thực \(a_1,a_2,a_3,....,a_n\left(n\in N^{\circledast}\right)\), ta có :
\(\left|a_1+a_2+...+a_n\right|\le\left|a_a\right|+\left|a_1\right|+....+\left|a_n\right|\)
Cho: \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}\) với \(a_1+a_2+...+a_n\)# 0. Tính:
1. A = \(\frac{a^2_1+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)^2}\)
2. B = \(\frac{a^9_1+a^9_2+...+a^9_n}{\left(a_1+a_2+...+a_n\right)^9}\)
Cho \(a_1;a_2;...a_n\ge0\) và \(a_1.a_2.a_3...a_n=1\)
CMR : \(\left(1+a_1\right)\left(1+a_2\right)+...+\left(1+a_n\right)\ge2\)
CM :\(\left(1+a_1\right)+\left(1+a_2\right)+...+\left(1+a_n\right)\ge2^n\)
Áp dụng BĐT Cô si cho 2 số \(a_1\) và 1 :
\(a_1+1\ge2\sqrt{a_1}\ge0\)
Tương tự cũng có :
\(a_2+1\ge2\sqrt{a_2}\ge0\)
........
\(a_n+1\ge2\sqrt{a_n}\ge0\)
=> \(\left(1+a_1\right)+\left(1+a_2\right)+...+\left(1+a_n\right)\ge2^n\sqrt{a_1.a_2...a_n}=2^n\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a_1=a_2=...=a_n=1\)
Mik sửa lại đề thành \(\left(1+a_1\right)+\left(1+a_2\right)+...+\left(1+a_n\right)\ge2^n\)
Cho \(n\) số \(a_1,a_2,...,a_n\in\left[0;1\right]\)
CMR:\(\left(1+a_1+a_2+a_3+...+a_n\right)^2\ge4\left(a^2_1+a^2_2+a^2_3+...+a^2_n\right)\)
Do \(a_1;a_2;...a_n\in\left[0;1\right]\Rightarrow\left\{{}\begin{matrix}0\le a_1\le1\\0\le a_2\le1\\...\\0\le a_n\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a_1\left(1-a_1\right)\ge0\\a_2\left(1-a_2\right)\ge0\\...\\a_n\left(1-a_n\right)\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a_1\ge a_1^2\\a_2\ge a_2^2\\...\\a_n\ge a_n^2\end{matrix}\right.\)
\(\Rightarrow a_1^2+a_2^2+...+a_n^2\le a_1+a_2+...+a_n\)
Do đó ta chỉ cần chứng minh:
\(\left(1+a_1+a_2+...+a_n\right)^2\ge4\left(a_1+a_2+...+a_n\right)\)
\(\Leftrightarrow1+2\left(a_1+a_2+...+a_n\right)+\left(a_1+a_2+...+a_n\right)^2\ge4\left(a_1+a_2+...+a_n\right)\)
\(\Leftrightarrow\left(a_1+a_2+...+a_n\right)^2-2\left(a_1+a_2+...+a_n\right)+1\ge0\)
\(\Leftrightarrow\left(a_1+a_2+...+a_n-1\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra tại \(\left(a_1,a_2,...,a_n\right)=\left(0,0,..,1\right)\) và các hoán vị
cho n số thực dương \(a_{_{ }1},a_2,...,a_n\)có tổng bằng 1. Chứng minh rằng:
a) \(\left(a_1+\frac{1}{a_2}\right)^2+\left(a_2+\frac{1}{a_3}\right)^2+...+\left(a_n+\frac{1}{a_1}\right)^2\ge\left(\frac{n^2+1}{n}\right)^2\)
b) \(\left(a_1+\frac{1}{a_1}\right)^2+\left(a_2+\frac{1}{a_2}\right)^2+...+\left(a_n+\frac{1}{a_n}\right)^2\ge\left(\frac{n^2+1}{n}\right)^2\)
Cho \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}\)
Tính:
a) \(\frac{a_1^2+a_2^2+...+a_n^2}{\left(a_1+a_2+...+a_n\right)^2}\) b) \(\frac{a_1^7+a_2^7+...+a_n^7}{\left(a_1+a_2+...+a_n\right)^7}\)
Help me, please!
Chả biết đúng hay sai! Cứ làm vậy
Ta có: \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}\)
\(=\frac{a_1+a_2+...+a_{n-1}+a_n}{a_2+a_3+..+a_n+a_1}=1\Rightarrow a_1=a_2=...=a_n\) (theo t/c tỉ dãy số bằng nhau)
Do đó:
a) \(\frac{a_1^2+a_2^2+...+a_n^2}{\left(a_1+a_2+...+a_n\right)^2}=\frac{na_1^2}{\left(na_1\right)^2}=\frac{na_1^2}{n^2a_1^2}=\frac{1}{n}\)
b) \(\frac{a_1^7+a_2^7+...+a_n^7}{\left(a_1+a_2+...+a_n\right)^7}=\frac{na_1^7}{\left(na_1\right)^7}=\frac{na_1^7}{n^7a_1^7}=\frac{n}{n^7}\)
Bạn gì có nhãn "CTV" gì ấy trả lời đúng không vậy mn? Đang bí bài này...=((