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ta có
a1+(a2+a3+a4)+... +(a11+a12+a13)+a14+(a15+a16+a17)+(a18+a19+a20)<0
a1>0; a2+a3+a4>0;...;a11+a12+a13>0;a15+a16+a17>0;a18+a19+a20>0; a14<0
Ta có:
(a1+a2+a3)+...+(a10+a11+a12)+(a13+a14)+(a15+a16+a17)+(a18+a19+a20)<0
=>(a13+a14)<0
có a12+a13+a14>0=>a12>0
Từ các cmt suy ra a1>0; a12>0; a14<0
=>a1. a14+a12.a12<a1.a12(đpcm)
# HOK TỐT #
ta có
a1+(a2+a3+a4)+... +(a11+a12+a13)+a14+(a15+a16+a17)+(a18+a19+a20)<0
a1>0; a2+a3+a4>0;...;a11+a12+a13>0;a15+a16+a17>0;a18+a19+a20>0; a14<0
Ta có:
(a1+a2+a3)+...+(a10+a11+a12)+(a13+a14)+(a15+a16+a17)+(a18+a19+a20)<0
=>(a13+a14)<0
có a12+a13+a14>0=>a12>0
Từ các cmt suy ra a1>0; a12>0; a14<0
=>a1. a14+a12.a12<a1.a12
Ta có : \(a_1+(a_2+a_3+a_4)+...+(a_{11}+a_{12}+a_{13})+a_{14}+(a_{15}+a_{16}+a_{17})+(a_{18}+a_{19}+a_{20})< 0\)
\(a_1>0;a_2+a_3+a_4>0;....;a_{11}+a_{12}+a_{13}>0;a_{15}+a_{16}+a_{17}>0;a_{18}+a_{19}+a_{20}>0\Rightarrow a_{14}< 0\)
Cũng như vậy : \((a_1+a_2+a_3)+...+(a_{10}+a_{11}+a_{12})+(a_{13}+a_{14})+(a_{15}+a_{16}+a_{17})+(a_{18}+a_{19}+a_{20})< 0\)
\(\Rightarrow a_{13}+a_{14}< 0\)
Mặt khác : \(a_{12}+a_{13}+a_{14}>0\Rightarrow a_{12}>0\)
Từ các điều kiện \(a_1>0;a_{12}>0;a_{14}< 0\Rightarrow a_1\cdot a_{14}+a_{14}\cdot a_{12}< a_1\cdot a_{12}(đpcm)\)
P/S : Hoq chắc :>
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{100}}{a_1}=\frac{a_1+a_2+...+a_{100}}{a_1+a_2+...+a_{100}}=1\)\(\Rightarrow\)\(a_1=a_2=...=a_{100}\)
\(\Rightarrow\)\(M=\frac{a_1^2+a_2^2+a_3^2+...+a_{100}^2}{\left(a_1+a_2+a_3+...+a_{100}\right)^2}=\frac{100a_1^2}{100^2a_1^2}=\frac{1}{100}\)
Ta thấy : \(a_1+a_2+a_3+.....+a_{2015}+a_1=1008.1=1008\)
Mà \(a_1+a_2+a_3+......+a_{2015}=0\)
\(\Rightarrow a_1+\left(a_1+a_2+a_3+....+a_{2015}\right)=1008\Leftrightarrow a_1+0=1008\) \(\Rightarrow a_1=1008\)
\(a_1+a_2+a_3+..+a_{2015}=0\)\(0\)
\(\Rightarrow\left(a_1+a_2\right)+...+\left(a_1+a_{2015}\right)\)\(=\frac{\left(2015-1\right)}{2}+1=1008\)
\(\Rightarrow a_1+\left(a_1+a_2+..+a_{2015}\right)=1008\)
\(\Rightarrow a_1=1008\)
Ta có:
\(a_1+a_2+...+a_{2015}=0\)
\(\Leftrightarrow\left(a_1+a_2\right)+\left(a_3+a_4\right)+...+\left(a_{2013}+a_{2014}\right)+\left(a_{2015}+a_1\right)-a_1=0\)
\(\Leftrightarrow1+1+...+1-a_1=0\)
\(\Leftrightarrow1008-a_1=0\)
\(\Leftrightarrow a_1=1008\)
Có: \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=.....=\frac{a_{2008}}{a_{2009}}=\frac{a_1+a_2+a_3+...+a_{2008}}{a_2+a_3+a_4+....+a_{2009}}\)(tính chất dãy tỉ số bằng nhau)
=> \(\left(\frac{a_1}{a_2}\right)^{2008}=\left(\frac{a_2}{a_3}\right)^{2008}=...=\left(\frac{a_{2008}}{a_{2009}}\right)^{2008}=\left(\frac{a_1+a_2+...+a_{2008}}{a_2+a_3+...+a_{2009}}\right)^{2008}\)
\(=\frac{a_1.a_2.....a_{2008}}{a_2.a_3.....a_{2009}}=\frac{a_1}{a_{2009}}\)
=> \(\frac{a_1}{a_{2009}}=\left(\frac{a_1+a_2+...+a_{2008}}{a_2+a_3+....+a_{2009}}\right)^{2008}\)
=> Đpcm
Ta có:
\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a2008}{a2009}=\frac{\left(a1+a2+...+a2008\right)}{\left(a2+a3+...+a2009\right)}\)
\(\Rightarrow\left(\frac{a1}{a2}\right)^{2008}=\left(\frac{a2}{a3}\right)^{2008}=..=\left(\frac{a2008}{a2009}\right)^{2008}=\left(\frac{a1+a2+..+a2008}{a2+a3+..+a2009}\right)^{2008}\)
\(\Rightarrow\frac{a1.a2....a2008}{a2.a3...a2009}=\left(\frac{a1+a2+..+a2008}{a2+a3+..+a2009}\right)^{2008}\)
\(\Rightarrow\frac{a1}{a2009}=\left(\frac{a1+a2+..+a2008}{a2+a3+..+a2009}\right)^{2008}\)