Ch 2010 số thực a1,a2,a3,..,a2010 thỏa mãn điều kiện
\(\hept{\begin{cases}a_1+a_2+...+a_{2010}=0\\a_1^2+a_2^2+...+a_{2010}^2=1\end{cases}}\)
chứng minh rằng trong 2010 số trên,có 2 số có tích không vượt quá -1/2010
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Ta có:
\(a_2^2=a_1.a_3;a_3^2=a_2.a_4;...;a^2_{2010}=a_{2009}.a_{2011}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3};\frac{a_2}{a_3}=\frac{a_3}{a_4};...;\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=...=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}=\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}\) (1)
Ta lại có:
\(\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}...\frac{a_{2009}}{a_{2010}}.\frac{a_{2010}}{a_{2011}}=\frac{a_1}{a_{2011}}\) (2)
Từ (1) và (2) ta suy ra
\(\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\)
Ta có :
\(a_2^2=a_1.a_3\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}\)
\(a^2_3=a_2.a_4\Rightarrow\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(............\)
\(a^2_{2010}=a_{2009}.a_{2011}\Rightarrow\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=........=\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=.......=\frac{a_{2010}}{a_{2011}}=k\)
\(\Rightarrow a_1=a_2.k\)
\(\Rightarrow a_1=a_3.k^2\)
\(\Rightarrow a_1=a_4.k^3\)
\(...............\)
\(\Rightarrow a_1=a_{2011}.k^{2010}\)
\(\Rightarrow\frac{a_1}{a_{2011}}=k^{2010}\) (1)
Ta có : \(k^{2010}=\left(\frac{a_1}{a_2}\right)^{2010}=\left(\frac{a_2}{a_3}\right)^{2010}=...=\left(\frac{a_{2010}}{a_{2011}}\right)^{2010}=\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=....=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}\)
\(=\frac{a_1^{2010}+a_2^{2010}+a_3^{2010}+....+a^{2010}_{2010}}{a_2^{2010}+a_3^{2010}+a_4^{2010}+....+a_{2011}^{2010}}\) ( theo TC DTSBN ) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a_1^{2010}+a_2^{2010}+....+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+....+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\) (đpcm)
Ta có :
\(a_2^2=a_1.a_3\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}\)
\(a_3^2=a_2.a_4\Rightarrow\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(.........\)
\(a_{2010}^2=a_{2009}.a_{2011}\Rightarrow\frac{a_{2019}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=.....=\frac{a_{2010}}{a_{2011}}=k\) ( k thuộc Z )
\(\Rightarrow a_1=a_2.k\)
\(\Rightarrow a_1=a_3.k_2\)
\(.........\)
\(\Rightarrow a_1=a_{2011}.k_{2010}\)
\(\Rightarrow\frac{a_1}{a_{2011}}=k^{2010}=\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=...=\frac{a_{2010}}{a_{2011}}=\frac{a^{2010}_1+a^{2010}_2+....+a_{2010}^{2010}}{a^{2010}_2+a^{2010}_3+....+a_{2011}^{2010}}\) (đpcm)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{2009}}{a_{2010}}=\frac{a_1+a_2+...+a_{2009}}{a_2+a_3+...+a_{2010}}\)
\(\Rightarrow\)\(\frac{a_1}{a_2}=\frac{a_1+a_2+...+a_{2009}}{a_2+a_3+...+a_{2010}}\)
\(\Rightarrow\)\(\left(\frac{a_1}{a_2}\right)^{2009}=\left(\frac{a_1+a_2+...+a_{2009}}{a_2+a_3+...+a_{2010}}\right)^{2009}\) \(\left(1\right)\)
Lại có :
\(\left(\frac{a_1}{a_2}\right)^{2009}=\frac{a_1}{a_2}.\frac{a_1}{a_2}.....\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}.....\frac{a_{2009}}{a_{2010}}=\frac{a_1.a_2.....a_{2009}}{a_2.a_3.....a_{2010}}=\frac{a_1}{a_{2010}}\) \(\left(2\right)\)
Từ (1) và (2) suy ra đpcm : \(\frac{a_1}{a_{2010}}=\left(\frac{a_1+a_2+...+a_{2009}}{a_2+a_3+...+a_{2010}}\right)^{2009}\) \(\left[=\left(\frac{a_1}{a_2}\right)^{2009}\right]\)
Chúc bạn học tốt ~
Ta có a1<a2<a3 => a1+a2+a3<3.a3
a4<a5<a6=> a4+a5+a6<3.a6
...........................................................
a2008<a2009<a2010=>a2008+a2009+a2010<3.a2010
=> a1+a2+a3+a4+a5+a6 +..+a2008+a2009+a2010 <3(a3+a6+...+a2010)
=>( a1+a2+a3+a4+a5+a6 +..+a2008+a2009+a2010 )/(a3+a6+...+a2010) <3