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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
Ta có : \(\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}< \dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}\left(1\right)\)
\(\sqrt{n}-\sqrt{n-1}=\dfrac{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n+1}\right)}{\sqrt{n}+\sqrt{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}>\dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)>\dfrac{1}{\sqrt{n}}\left(2\right)\)
Từ \(\left(1;2\right)\text{⇒ }đpcm\)
nè
a)
+) Ta có: \(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\) \(=\dfrac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+1-n}\)
\(=2\left(\sqrt{n+1}-\sqrt{n}\right)\) (1)
+) Ta có:
\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}< \dfrac{2}{\sqrt{n}+\sqrt{n-1}}=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\) \(=\dfrac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{n-\left(n-1\right)}\)
\(=2\left(\sqrt{n}-\sqrt{n-1}\right)\) (2)
Từ (1) và (2) ⇒ đpcm
Học toán vui vẻ! 
Lời giải
Với mọi $n\in\mathbb{N}$ ta có:
\(\frac{1}{\sqrt{1}}> \frac{1}{\sqrt{n}}\)
\(\frac{1}{\sqrt{2}}> \frac{1}{\sqrt{n}}\)
.....
Do đó:
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}> \underbrace{\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}+...+\frac{1}{\sqrt{n}}}_{\text{n số}}=\frac{n}{\sqrt{n}}=\sqrt{n}\)
(chứng minh xong vế 1)
Vế 2:
\(\frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+...+\frac{1}{2\sqrt{n}}< \frac{1}{\sqrt{0}+\sqrt{1}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}\)
\(=\frac{\sqrt{1}-\sqrt{0}}{1-0}+\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n}-\sqrt{n-1}}{n-(n-1)}\)
\(=\sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+...+\sqrt{n}-\sqrt{n-1}=\sqrt{n}\)
\(\Rightarrow \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}< 2\sqrt{n}\) (đpcm)
Vậy....