Proof of $AM \geq GM$ using integration

We like to prove :

Observe that for any $x > 0$:

because, so long as the integrand’s $t$  lies strictly increasing or decreasing between $x$  and $G$, the sign $G – x$  and of $1/t – 1/G$, must be the same.

Equality sign holds iff $x = G$.

Now, replace $x$  by $x_i$ and sum up over $i = 1, 2, … , n$.

 $A \geq G$.