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\(a_n=\frac{1+\left(\frac{n}{n+2}\right)^n}{1-\left(\frac{n}{n+2}\right)^n}\)
\(a_n=\frac{\left(\frac{n}{n+2}\right)^2-\left(-1\right)}{\left(1-\frac{n}{n+2}\right)\left(1+\frac{n}{n+2}\right)}\)
\(a_n=\frac{\left(\frac{n}{n+2}-1\right)\left(\frac{n}{n+2}+1\right)}{\left(1-\frac{n}{n+2}\right)\left(1+\frac{n}{n+2}\right)}\)
\(a_n=\frac{\left(\frac{n}{n+2}-1\right)}{\left(1-\frac{n}{n+2}\right)}\)
\(a_n=\frac{-\left(1-\frac{n}{n+2}\right)}{\left(1-\frac{n}{n+2}\right)}\)
\(a_n=1\)
\(\Rightarrow\hept{\begin{cases}a=1\\n=1\end{cases}}\)
vậy \(\hept{\begin{cases}a=1\\n=1\end{cases}}\)
\(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\left(1-\frac{2}{2.3}\right)\left(...\right).....\left[1-\frac{2}{n\left(n+1\right)}\right]=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{\left(n-2\right)\left(n+1\right)}{\left(n-1\right).n}.\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}=\)
\(=\frac{1}{3}.\frac{n+2}{n}=\frac{1}{3}-\frac{1}{3}.\frac{2}{n}>\frac{1}{3}\)
Áp dụng bất đẳng thức Cô - si với n số dương ta được
\(a_1+a_2+...+a_n\ge n\sqrt[n]{a_1.a_2....a_n}\)
\(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\ge n\sqrt[n]{\frac{1}{a_1}.\frac{1}{a_2}....\frac{1}{a_n}}\)
Suy ra \(\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\ge n^2.\sqrt[n]{1}=n^2\)
(dấu "=" xẩy ra <=> a1=a2 =...=an)
Theo bat dang thuc cauchy ta co
a1+a2+...+an lon hon hoc bang n.can bac n cua (a1.a2....an) (1)
1/a1+1/a2...1/an lon hon hoac bang n.1/can bac n cua (a1.a2...an) (2)
Nhan 2 ve (1) va (2) ta duoc
(a1+a2+...+an).(1/a1+1/a2+...1/an) lon hon hoac bang n tren 2
=>1/a1+1/a2+...1/an lon hon hoac bang n tren 2/a1+a2+...+an
Dau bang xay ra khi a1=a2=...=an
Mk giai co hieu ko