Đặt \(A=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{n}}\)

\(A=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+\dfrac{2}{\sqrt{n}+\sqrt{n}}\)

\(A>2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)

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

\(A>2\left(\sqrt{n+1}-1\right)\)

Cần cm:\(2\left(\sqrt{n+1}-1\right)>\sqrt{n}\)

\(\Leftrightarrow4\left(n+1\right)+4-8\sqrt{n+1}>n\)

\(\Leftrightarrow3n+8>8\sqrt{n+1}\)

Lại có:\(8\sqrt{n+1}\le2\left(n+1\right)+8=2n+10\le3n+8\)(AM-GM)

Dấu "=" không xảy ra

=>đpcm

quỳnh đăng lên giúp ai zậy ns đi nghe xem nào

Câu hỏi của Cường Hoàng - Toán lớp 9 | Học trực tuyến

Áp dụng : \(\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n-1}}+...+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{2}}+1>2\left(\sqrt{n+1}-\sqrt{n}\right)+2\left(\sqrt{n}-\sqrt{n-1}\right)+...+2\left(\sqrt{4}-\sqrt{3}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+2\left(\sqrt{2}-1\right).\)

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

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

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

Do đó:

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

\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)

bai 1

(n+1)√n=√n^3+√n>2√(n^3.n)=2n^2>2(n^2-1)=2(n-1)(n+1)

1/[(n+1)√n]<1/[2(n-1)(n+1)]=1/4.[2/(n-1)(n+1)]

A=..

n =1 yes

n>1

A<1+1/4[2/1.3+2/3.5+..+2/(n-1)(n+1)

A<1+1/4[ 2-1/(n+1)]<1+1/2<2=>dpcm