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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 \(a=b=2\)
\(\frac{1}{\sqrt{1}+\sqrt{2}}+....\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\) (40 số)
................................................................\(>\frac{40}{10}=4\)
=>đpcm
hc tốt
ko chắc lắm :)
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)
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
\(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge2+\sqrt{22+\frac{1}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\ge2+\sqrt{22+\frac{\left(a+b+c\right)^3}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\ge2\sqrt{abc}+\sqrt{22abc+\left(a+b+c\right)^3}\)
\(\Leftrightarrow\left(a+b+c\right)^2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\)
\(\ge\left(a+b+c\right)^3+26abc+4\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\)
\(\Leftrightarrow\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge13abc+2\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\left(1\right)\)
Ta có:
\(VT_{\left(1\right)}=\frac{13}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\frac{14}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge13abc+\frac{14}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
Ta cần chứng minh:
\(\frac{14}{27}\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge2\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\)
\(\Leftrightarrow7\left(a+b+c\right)^2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge27\sqrt{abc}\sqrt{\left(a+b+c\right)^3+22abc}\)
\(\Leftrightarrow49\left(a+b+c\right)^4\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)^2\)
\(\ge27^2abc\left[\left(a+b+c\right)^3+22abc\right]\left(2\right)\)
Lại có:
\(VT_{\left(2\right)}\ge49\left(a+b+c\right)^3\cdot27abc\)
Ta chứng minh
\(49\left(a+b+c\right)^3\cdot27abc\ge27^2abc\left[\left(a+b+c\right)^3+22abc\right]\)
\(\Leftrightarrow\left(a+b+c\right)^3\ge27abc\)
Bất đẳng thức cuối cùng hiển nhiên đúng theo Cô-si 3 số.
Vậy bđt đã được chứng minh.
Dấu = khi a=b=c
Ta có
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)
Áp dụng vào A ta được
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{80}-\sqrt{79}\)
\(=\sqrt{80}-1>\sqrt{25}-1=4\)
Chỗ nào không hiểu thì cứ hỏi nhé