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\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}+\frac{\sqrt{n+1}}{n+1}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{99}}{99}-\frac{\sqrt{100}}{100}\)
\(=1-\frac{\sqrt{100}}{100}=\frac{9}{10}< 1\)
Đặt \(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)
Ta có: \(\frac{1}{1+\sqrt{2}}>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(\frac{1}{\sqrt{3}+\sqrt{4}}>\frac{1}{2}\left(\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}\right)\)
...
\(\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{2}\left(\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
Cộng các bất đẳng thức trên lại với nhau, ta được:
\(A>\frac{1}{2}\left(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{81}-\sqrt{80}}{81-80}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{81}-\sqrt{80}\right)\)
\(\Leftrightarrow A>\frac{1}{2}\left(\sqrt{81}-1\right)=\frac{1}{2}\cdot\left(9-1\right)=\frac{1}{2}\cdot8=4\)
\(\Leftrightarrow A>4\)(đpcm)
Lời giải:
Sửa đề: CMR:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\sqrt{n}-1\)
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Sử dụng PP liên hợp ta có:
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{n-1}+\sqrt{n}}=\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{3}+\sqrt{4})(\sqrt{4}-\sqrt{3})}+....+\frac{\sqrt{n}-\sqrt{n-1}}{(\sqrt{n-1}+\sqrt{n})(\sqrt{n}-\sqrt{n-1})}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+....+\frac{\sqrt{n}-\sqrt{n-1}}{n-(n-1)}\)
\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{n}-\sqrt{n-1}\)
\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)
Ta có đpcm.
+ \(2\cdot\frac{1}{\sqrt{n}+\sqrt{n+1}}< \frac{2}{\sqrt{n}+\sqrt{n}}< 2\cdot\frac{1}{\sqrt{n-1}+\sqrt{n}}\) \(\Rightarrow2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)
\(\Rightarrow A>2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{101}-\sqrt{100}\right)\)
\(\Rightarrow A>2\left(\sqrt{101}-\sqrt{2}\right)>17\)
+ \(A< 2\left(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\Rightarrow A< 2\left(\sqrt{100}-1\right)=18\)
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 \(a=b=2\)