An alternative is the use of polar coordinates...
(1)
... so that is...
(2)
Now the proof of the 'inequality' is equivalent to prove that is...
(3)
... and (3) is true because the function has its maximum for and is ...
Kind regards
Dear friends,
while working on some computations, I figured out that my arguments supply a proof for
the arithmetic and geometric mean inequality by using calculus. I wanted to share here.
Assertion. For all show that with equality if and only if .
Proof. The case is trivial.
Let and define
.
We shall prove that for all , with equality if and only if .
In this case, we may find such that (in particular, ), which yields
, where for .
As , we have to show that for all , with equality if and only if .
We can compute that
and for all .
The function is therefore convex and this shows that is the global minimum value
since . Finally, for all , with equality if and only if .