For non-negative and ,
with equality exactly when all the are equal.
Lemma. For real , there exist real with the same symmetric averages .
Proof. We consider the derivative of . The roots of are . Without loss of generality, we assume that the increase as increases. Now for any , must have a root between and by Rolle's theorem if , and if , then is a root of times, so it must be a root of times. It follows that must have non-positive, real roots, i.e., for some non-negative reals ,
It follows that the symmetric sum for is , so the symmetric average .
Thus to prove Newton's theorem, it is sufficient to prove
Expanding the left side, we see that this is
But this is clearly equivalent to
which holds by the rearrangement inequality.
Proof: without calculus
We will proceed by induction on .
For , the inequality just reduces to AM-GM inequality. Now suppose that for some positive integer the inequality holds.
Let , , , be non-negative numbers and be the symmetric averages of them. Let be the symmetric averages of , , . Note that .
By induction this completes the proof.