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\(1-\frac{2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(\left(1-\frac{2}{2.3}\right)\left(...\right).....\left[1-\frac{2}{n\left(n+1\right)}\right]=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}....\frac{\left(n-2\right)\left(n+1\right)}{\left(n-1\right).n}.\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}=\)
\(=\frac{1}{3}.\frac{n+2}{n}=\frac{1}{3}-\frac{1}{3}.\frac{2}{n}>\frac{1}{3}\)
Áp dụng bđt sau : \(\frac{a^n+b^n}{2}\ge\frac{\left(a+b\right)^n}{2}\)ta được
\(\frac{1}{\left(1+a\right)^n}+\frac{1}{\left(1+b\right)^n}\ge2\left(\frac{\frac{1}{1+a}+\frac{1}{1+b}}{2}\right)^n\)
Ta đi c/m bđt phụ : Với a,b > 1 thì \(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\)(1)
Bđt (1) \(\Leftrightarrow\frac{\left(a+b\right)+2}{1+\left(a+b\right)+ab}\ge\frac{2}{1+\sqrt{ab}}\)(Quy đồng VT)
\(\Leftrightarrow\left(a+b\right)+2+\left(a+b\right)\sqrt{ab}+2\sqrt{ab}\ge2+2\left(a+b\right)+2ab\)
\(\Leftrightarrow\left(a+b\right)\left(\sqrt{ab}-1\right)+2\sqrt{ab}\left(1-\sqrt{ab}\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{ab}-1\right)\left(a+b-2\sqrt{ab}\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{ab}-1\right)\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)(Luôn đúng vs mọi a;b > 1)
Áp dụng bđt (1) được
\(\frac{1}{\left(1+a\right)^n}+\frac{1}{\left(1+b\right)^n}\ge2\left(\frac{\frac{1}{1+a}+\frac{1}{1+b}}{2}\right)^n\ge2\left(\frac{1}{1+\sqrt{ab}}\right)^n=\frac{2}{\left(1+\sqrt{ab}\right)^n}\)
Dấu "=" xảy ra tại a = b
Áp dụng buổi thức đơn ta được
\(\sqrt[a]{b}\)\(a+b:2\)\(>\)ta được
\(\frac{1}{1+A}\)+ \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(\frac{A+B=2}{ }\)
\(\frac{A+B=2}{1+A+B}\)
\(VẬY\)Nếu bạn làm tắt theo mik thì
Mik chưa ra đáp án được vì
\(B\sqrt[A]{B}\)CHỖ B BỊ LỖI
MAGICPENCIL,HÃY LUÔN :-)
\(\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{\left(n-1\right)^2-1}{\left(n-1\right)^2}\right)\left(\frac{n^2-1}{n^2}\right)\)
=\(\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}.\frac{\left(4-1\right)\left(4+1\right)}{4^2}...\frac{\left(n-2\right)n}{\left(n-1\right)^2}.\frac{\left(n-1\right)\left(n+1\right)}{n^2}\)
=\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{\left(n-2\right).n}{\left(n-1\right)^2}.\frac{\left(n-1\right)\left(n+1\right)}{n^2}=\frac{1}{2}.\frac{n+1}{n}=\frac{1}{2}+\frac{1}{2n}>\frac{1}{2}\)
Cách lớp 7 nà:)
\(\frac{1}{n.\left(n+1\right)^2}=\frac{1}{n.\left(n+1\right).\left(n+1\right)}< \frac{1}{n.n\left(n+1\right)}< \frac{1}{\left(n-1\right)n\left(n+1\right)}\) (n>=2_
\(\text{Suy ra }VT< \frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
Mặt khác ta có công thức \(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]}{2}\) (n>= 2)
Suy ra \(VT< \frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}\right)< \frac{1}{2}.\frac{1}{2}=\frac{1}{4}\left(\text{do }\frac{1}{n\left(n+1\right)}>0\right)\)
Vậy ta có đpcm
Gắt chưa??? :>> Dương Bá Gia Bảo