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\(\frac{1}{n\sqrt{n-1}}=\frac{\sqrt{n-1}}{\left(n-1\right)n}=\sqrt{n-1}.\frac{1}{\left(n-1\right)n}=\sqrt{n-1}\left(\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\sqrt{n-1}\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{n}}\right)\)
\(=\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\frac{\sqrt{n-1}}{\sqrt{n}}\right)\)
\(< \left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right)\)
Áp dụng vài bài toán:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)
\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+2\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+2\left(\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2004}}-\frac{1}{\sqrt{2005}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2005}}\right)=2-\frac{2}{\sqrt{2005}}< 2\)
Vậy \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}< 2\)
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
Mình đã chứng minh \(\frac{1}{2\sqrt{n+1}}< \sqrt{n+1}-\sqrt{n}\left(n\inℕ^∗\right)\) rồi nha!
Áp dụng vào, ta được: \(\frac{1}{2\sqrt{1}}< \sqrt{1}\)
\(\frac{1}{2\sqrt{2}}< \sqrt{2}-\sqrt{1}\)
\(\frac{1}{2\sqrt{3}}< \sqrt{3}-\sqrt{2}\)
.............................
\(\frac{1}{2\sqrt{2500}}< \sqrt{2500}-\sqrt{2499}\)
\(\Rightarrow1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2500}}\)
\(< 2\left(\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2500}-\sqrt{2499}\right)\)
\(=2.50=100\)
=> ĐPCM
P/s: sai sót xin bỏ qua cho.
Với mọi n nguyên dương ta có:
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=1\Rightarrow\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)
Với k nguyên dương thì
\(\frac{1}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k+1}+\sqrt{k}}\Rightarrow\frac{2}{\sqrt{k-1}+\sqrt{k}}>\frac{1}{\sqrt{k-1}+\sqrt{k}}+\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k}-\sqrt{k-1}+\sqrt{k+1}-\sqrt{k}\)
\(=\sqrt{k+1}-\sqrt{k-1}\)(*)
Đặt A = vế trái. Áp dụng (*) ta có:
\(\frac{2}{\sqrt{1}+\sqrt{2}}>\sqrt{3}-\sqrt{1}\)
\(\frac{2}{\sqrt{3}+\sqrt{4}}>\sqrt{5}-\sqrt{3}\)
...
\(\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-\sqrt{79}\)
Cộng tất cả lại
\(2A=\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{3}+\sqrt{4}}+....+\frac{2}{\sqrt{79}+\sqrt{80}}>\sqrt{81}-1=8\Rightarrow A>4\left(đpcm\right)\)
3.
Theo bất đẳng thức cô si ta có:
\(\sqrt{b-1}=\sqrt{1.\left(b-1\right)}\le\frac{1+b-1}{2}=\frac{b}{2}\Rightarrow a.\sqrt{b-1}\le\frac{a.b}{2}\)
Tương tự \(\Rightarrow b.\sqrt{a-1}\le\frac{a.b}{2}\Rightarrow a.\sqrt{b-1}+b.\sqrt{a-1}\le a.b\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=2\)
\(a+b+c=abc\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
\(\sqrt{1+\frac{1}{a^2}}+\sqrt{1+\frac{1}{b^2}}+\sqrt{1+\frac{1}{c^2}}\ge\sqrt{\left(1+1+1\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(\ge\sqrt{3^2+3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}=\sqrt{9+3}=\sqrt{12}=2\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c=\(\sqrt{3}\)
dễ thế mà ko làm được
\(\sqrt{1+\frac{\sqrt{3}}{2}}=\frac{1}{2}\sqrt{4+2\sqrt{3}}\)
\(=\frac{1}{2}\sqrt{\left(1+\sqrt{3}\right)^2}=\frac{\sqrt{3}+1}{2}\)