Question

CMR : \(\sqrt{n}\)< \(\dfrac{1}{\sqrt{1}}\)+\(\dfrac{1}{\sqrt{2}}\)+\(\dfrac{1}{\sqrt{3}}\)+...+\(\dfrac{1}{\sqrt{n}}\) với n ≥2; n ϵ Z⁺

Answer 1

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