Question on the inequality Question.

prove  that if ${ a }_{ 1 },{ a }_{ 2 },...{ a }_{ n }>0$ then $$ \frac { { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } }  } $$ 

Proof 
 $$\\ \left( { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } \right) \left( \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } }  \right) =\underset { n\left( n-1 \right) /2\quad terms }{ \underbrace { \left( \frac { { a }_{ 1 } }{ { a }_{ 2 } } +\frac { { a }_{ 2 } }{ { a }_{ 1 } }  \right) +...+\left( \frac { { a }_{ n-1 } }{ { a }_{ n } } +\frac { { a }_{ n } }{ { a }_{ n-1 } }  \right) + } \quad n\ge  } \\ \ge n+2\cdot \frac { n\left( n-1 \right)  }{ 2 } ={ n }^{ 2 }$$
in this  proof i didn't understand  this step "$n\left( n-1 \right) /2\quad ?
terms$" I mean how the number of terms can be $n\left( n-1 \right) /2\quad $ .Can anybody explain it.
Thanks in advance!

You have $a_i\cdot \frac1{a_i}$ for each $i$; these sum up to $n$.
Apart from that,  you have for each of the $n\choose 2$ choices of $i<j$ the summands $\frac{a_i}{a_j}+\frac{a_j}{a_i}$, which is $\ge 2$
