For , we define the symmetric sum to be the coefficient of in the polynomial (see Viete's sums). We define the symmetric average to be .
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
for any . Since this is a homogenous inequality, we may normalize it so that . The inequality then becomes
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.