Proof of AM≥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 $i=1,2,\dots ,n$.

 $A\ge G$.