\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n}+\sqrt{n+1}}\ge2014\)

\(\Rightarrow\frac{1-\sqrt{2}}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+...+\frac{\sqrt{n}-\sqrt{n+1}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n}-\sqrt{n+1}\right)}\)

\(=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+...+\frac{\sqrt{n}-\sqrt{n+1}}{n-\left(n+1\right)}\)

\(=\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{n}-\sqrt{n+1}}{-1}\)

\(=\frac{1-\sqrt{n+1}}{-1}=\sqrt{n+1}-1\ge2014\)

                                  \(\Leftrightarrow\sqrt{n+1}\ge2015\)

                                 \(\Leftrightarrow n+1=2015^2=4060225\)

\(V~~n=4060224\)

a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)

\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)

\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)

\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)

\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)

=>đpcm

b) Từ công thức trên ta có:

\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)

=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)

Ta có:

\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)

\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)

\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)

Đật 3 cái mẫu bên VT lần lượt là x,y,z rồi áp dụng C-S dạng engel

Để dễ nhìn ta đặt \(\hept{\begin{cases}\sqrt{2x-3}=a\\\sqrt{y-2}=b\\\sqrt{3z-1}=c\end{cases}\left(a,b,c\ge0\right)}\)

Vậy BĐT đầu tương đương \(T=\frac{1}{a}+\frac{4}{b}+\frac{16}{c}+a+b+c\)

Áp dụng BĐT C-S dạng Engel ta có:

\(\frac{1}{a}+\frac{4}{b}+\frac{16}{c}=\frac{1^2}{a}+\frac{2^2}{b}+\frac{4^2}{c}\ge\frac{\left(1+2+4\right)^2}{a+b+c}=\frac{49}{a+b+c}\)

Tiếp tục dùng AM-GM ta có: \(VT\ge\frac{49}{a+b+c}+\left(a+b+c\right)\ge2\sqrt{\frac{49}{a+b+c}\cdot\left(a+b+c\right)}=2\sqrt{49}=14\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a=1\\b=2\\c=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=\frac{17}{3}\end{cases}}\)

nhìn qua thì chắc AM-GM+Cauchy-schwarz chắc thế :)

\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{n^2\left(n+1\right)^2+n^2+\left(n+1\right)^2}{n^2\left(n+1\right)^2}=\frac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n\left(n+1\right)^2}=\frac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}\)

\(\Rightarrow p=n+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}+\frac{101}{n+1}\)

\(p=n+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}+\frac{101}{n+1}\)

\(p=n+1-\frac{1}{n+1}+\frac{101}{n+1}=n+1+\frac{100}{n+1}\ge2\sqrt{\frac{100\left(n+1\right)}{n+1}}=20\)

\(p_{min}=20\) khi \(n+1=\frac{100}{n+1}\Leftrightarrow n=9\)

bạn giải thích cho mình chỗ dấu suy ra thứ 2 được không ạ, vì sao lại xuất hiện n+1/1.2 +......... vậy ạ?

Nguyễn Hồng Nhung

Thay vào công thức:

\(\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}=1+\frac{1}{1.2}\) ; \(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2.3}\) ...

Cộng lại:

\(1+\frac{1}{1.2}+1+\frac{1}{2.3}+...+1+\frac{1}{n\left(n+1\right)}\)

Có n số 1 cộng với nhau ra n

CÒn lại đống \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}\) thôi

Ta có :

\(\frac{1}{\sqrt{k}}=\frac{2}{2\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}\)

\(=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\)

\(=2\left(\sqrt{k+1}-\sqrt{k}\right)\)

Vậy : \(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{n}}>2\left(\sqrt{2}-1\right)+2\left(\sqrt{3}-\sqrt{2}\right)+....+2\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(=2\left(\sqrt{n+1}-1\right)\left(đpcm\right)\)

Xét hạng tổng quát:

\(\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{1}{\sqrt{n}+\sqrt{n-1}}=\frac{\sqrt{n}-\sqrt{n-1}}{n-n+1}=\sqrt{n}-\sqrt{n-1}\)

Áp dụng vào bài, ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)

\(=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(=\sqrt{n}-1\)