$A_{n}$ is the arithmetic mean. With $a_{1} < A_{n}$ and $a_{2} > A_{n}$, and $\bar a_{1} = A_{n} $and $ \bar a_{2} = a_{1} + a_{2} - \bar a_{1}$, prove $\bar a_{1}\bar a_{2} \geq a_{1}a_{2}$
I'm having some trouble with this question, even though I feel like it is supposed to be super simple. Can I say that $A_{n} = (a_{1} + a_{2})/2$?
$$a_1<A_n<a_2\iff a_2-A_n>0, \ A_n - a_1 >0\\ \iff (a_2-A_n)(A_n-a_1)>0 \\ \iff (a_2+a_1)A_n - A_n^2 - a_2a_1 > 0 \\ \iff (a_2+a_1-A_n)A_n>a_2a_1\\ \iff \bar{a}_2\bar{a}_1 > a_2a_1 $$