Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(...........\)

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

Cộng theo vế ta có:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{100}{10}=10\) (Đpcm)

Ta có : \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}>\dfrac{1}{\sqrt{100}}.100=10\)

=>ĐPCM

Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(VT>\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}+\frac{1}{\sqrt{100}}\)

\(=\frac{1}{10}+\frac{1}{10}\) có 100 số hạng 

\(=\frac{100}{10}=10\)

Dòng 6 cuối cùng mình làm cũng không được chắc chắn lắm đâu òng 6 đấy bạn ngoặc ở dưới 1/10 +1/10 nhé

Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

.......

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=10\) (đpcm)

Ta có: \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}}\)

..........

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

Cộng các vế lại ta được:

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}.100=10\)

Vậy...

ta thấy:

\(\frac{1}{\sqrt{1}}>\frac{1}{10}\) 

\(\frac{1}{\sqrt{2}}>\frac{1}{10}\) 

   ....

\(\frac{1}{\sqrt{100}}=\frac{1}{10}\) 

=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}=\frac{1}{10}.100=10\) 

=> \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>10\) (đpcm)

Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}}=\frac{1}{10}\)

...

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}=\frac{1}{10}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>\frac{1}{10}+\frac{1}{10}+..+\frac{1}{10}=100.\frac{1}{10}=10\)Vậy \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}>10\left(đpcm\right)\)

Ta có : \(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{100}};\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{100}};\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{100}};...;\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{100}}\)

\(\Rightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>100.\frac{1}{\sqrt{100}}=\frac{100}{10}=10\)

\(10=\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}+\frac{1}{10}\)  (100 số hạng)

Ta có:

\(\frac{1}{\sqrt{1}}>\frac{1}{10}\);  \(\sqrt{2}< 10\Rightarrow\frac{1}{\sqrt{2}}>\frac{1}{10}\)....\(\frac{1}{\sqrt{99}}>\frac{1}{10}\)

Cộng vế theo vế 99 bđt trên ta được

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}>99\cdot\frac{1}{10}\)

\(\Leftrightarrow\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>100\cdot\frac{1}{10}=10\) (đpcm)

\(\sqrt{1}\frac{1}{\sqrt{100}}\)

\(\sqrt{2}\frac{1}{\sqrt{100}}\)

..................

\(\sqrt{99}\frac{1}{\sqrt{100}}\)

\(\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\)

Cộng từng vế của các bất đẳng thức trên ta được : 

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=\frac{100}{\sqrt{100}}=\frac{100}{10}=10\)

1/căn 1>1/10

1/ căn 2>1/10

...

1/căn 100>1/10

=>A>1/10.100=10