\(Cho\) \(a_1+a_2+...+a_n=1\) , \(n\in N\)
\(CMR:\)\(a^2_1+a^2_2+...+a^2_n\ge\frac{1}{n}\)
"Giúp mình với ạ!!"
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 các số \(a_1\cdot a_2\cdot...\cdot a_n>b>0\). CMR:
\(\frac{b}{b+a^2_1}+\frac{b}{b+a^2_2}+...+\frac{b}{b+a^2_n}\ge\frac{b\cdot n}{b+a_1\cdot a_2\cdot...\cdot a_n}\)
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 \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1};a_1+a_2+..+a_{n-1}+a_n\ne0\)
Tính \(\frac{a^2_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)^2}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=k\)
=>\(\frac{a_1}{a_2}.\frac{a_2}{a_3}.....\frac{a_{n-1}}{a_n}.\frac{a_n}{a_1}=k.k.....k.k\)
=>\(k^n=\frac{a_1.a_2.....a_{n-1}.a_n}{a_2.a_3.....a_n.a_1}\)
=>\(k^n=1=1^n\)
=>k=1
=>\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=1\)
=>\(a_1=a_2=...=a_n\)
\(=>\frac{a^2_1+a^2_2+...+a_n^2}{\left(a_1+a_2+...+a_n\right)^2}\)
=\(\frac{a^2_1+a^2_1+...+a_1^2}{\left(a_1+a_1+...+a_1\right)^2}\)
=\(\frac{n.a^2_1}{\left(n.a_1\right)^2}=\frac{n.a_1^2}{n^2.a^2_1}=\frac{1}{n}\)
thế này dc ko
Áp dụng t/c của dãy tỉ số bằng nhau, ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=\frac{a_1+a_2+...+a_{n-1}+a_n}{a_2+a_3+...+a_n+a_1}\Rightarrow a_1=a_2=...=a_n\)
\(\frac{a^1_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)}=\frac{na^2_1}{\left(na_1\right)^2}=\frac{1}{n}\)
Chứng minh rằng: \(\frac{a_1^2+a^2_2+...+a^2_n}{n}\ge\left(\frac{a_1+a_2+..+a_n}{n}\right)^2\)
Chiều mình ktra rồi, help me
Cho A=\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=.......=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}\)(a1 + a2 + a3 +.......+ an khác 0)
1.tính A=\(\frac{a_1^2+a^2_2+.......a^2_n}{\left(a_1+a_2+.......a_n\right)^2}\)
2) tính B=\(\frac{a_1^9+a^9_2+.......a^9_n}{\left(a_1+a_2+.......a_n\right)^9}\)
Giải hộ mình chi tiết nhất có thể nha. Mik sẽ tích và xin cảm
Chứng minh:
\(\left(a_1+a_2+...+a_n\right)^2\le n\left(a_1^2+a^2_2+...+a^2_n\right)\)
Ta có: \(\left\{{}\begin{matrix}a_1^2+a_2^2\ge2a_1a_2\\a_1^2+a_3^2\ge2a_1a_3\\...................\\a_{n-1}^2+a_n^2\ge2a_{n-1}a_n\end{matrix}\right.\)
\(\Rightarrow\left(n-1\right)\left(a_1^2+a_2^2+...+a_n^2\right)\ge2\left(a_1a_2+a_1a_3+...+a_{n-1}a_n\right)\)
\(\Leftrightarrow n\left(a_1^2+a_2^2+...+a_n^2\right)\ge2\left(a_1a_2+a_1a_3+...+a_{n-1}a_n\right)+\left(a_1^2+a_2^2+...+a_n^2\right)\)
\(\Leftrightarrow n\left(a_1^2+a_2^2+...+a_n^2\right)\ge\left(a_1+a_2+...+a_n\right)^2\)
Áp dụng BĐT căn trung bình bình phương ta có:
\(\sqrt{\dfrac{a_1^2+a_2^2+....+a^2_n}{n}}\ge\dfrac{a_1+a_2+...+a_n}{n}\)
\(\Leftrightarrow\dfrac{a_1^2+a_2^2+....+a^2_n}{n}\ge\left(\dfrac{a_1+a_2+...+a_n}{n}\right)^2\)
\(\Leftrightarrow\dfrac{a_1^2+a_2^2+....+a^2_n}{n}\ge\dfrac{\left(a_1+a_2+...+a_n\right)^2}{n^2}\)
\(\Leftrightarrow a_1^2+a_2^2+....+a^2_n\ge\dfrac{\left(a_1+a_2+...+a_n\right)^2}{n}\)
\(\Leftrightarrow n\left(a_1^2+a_2^2+....+a^2_n\right)\ge\left(a_1+a_2+...+a_n\right)^2\)
Khi \(a_1=a_2=...=a_n\)
Cho a1 , a2 , ... , an > 0 . Với mọi m,n ∈ N* , n ≥ 2 , chứng minh bất đẳng thức :
\(a_1^m+a_2^m+...a_n^m\ge\left(n-1\right)a_1a_2...a_n+\frac{a_1^{m-1}+a_2^{m-1}+...+a_n^{m-1}}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}\)
(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: \(Max\left\{A_1,A_2\right\}\ge\frac{n}{2}\)