The arithmetic mean of two unequal positive real numbers #p,q# is #(p+q)/2#, while the geometric mean is #sqrt(pq)#.
Thus, we need to first prove that #(p+q)/2>sqrt(pq)#. Multiply both sides by #2# and square both sides (since both sides are positive numbers, squaring both sides does not change the equality).
This gives #p^2+2pq+q^2>4pq#. Now, subtract both sides by #4pq# to get #p^2-2pq+q^2>0#.
The left-hand side can be factored to #(p-q)^2>0#, which is always true since a real number squared is always greater than or equal to #0#.
The second part requires us to prove that #sqrt(pq)>(sqrt(pq))^2/((p+q)/2)#.
We already have proved that #(p+q)/2>sqrt(pq)#, and both sides are positive. Find the reciprocal of both sides (which requires us to flip the inequality sign from #># to #<#).
This gives #1/((p+q)/2)< 1/sqrt(pq)#. Now, simply multiply both sides by #pq#.
This gives #(pq)/((p+q)/2)< sqrt(pq)#, or #(sqrt(pq))^2/((p+q)/2)< sqrt(pq)#, completing the proof.
