Vtyjkyuk

Definition of Newton's Inequalities: https://en.wikipedia.org/wiki/Newton%27s_inequalities

Everywhere I look it says equality holds iff all $a_1=...=a_n$, but let $n=3, a_1=0, a_2=0, a_3=1$, then $S_1=1/3, S_2=0, S_3=0$, then $S_1S_3=S_2^2$ equality holds.

What really is the equality case of Newton's Inequality?

For positive numbers $a_i$ the equality occurs for $a_1=a_2=...=a_n$.

It follows immediately from the proof.

$a_i$ are roots of the equation $$(x-a_i)(x-a_2)...(x-x_n)=0$$ or
$$x^n-binomn1S_1x^n-1+...+(-1)^nbinomnnS_n=0.$$
Thus, the equation
$$S_nx^n-binomn1S_n-1x^n-1+...+binomnn(-1)^n=0$$ has $n$ positive roots and by Rolle the equation
$$left(S_nx^n-binomn1S_n-1x^n-1+...+binomnn(-1)^nright)'=0$$ or
$$nS_nx^n-1-(n-1)binomn1S_n-1x^n-2+...+(-1)^n-1binomnn-1S_1=0$$
has $n-1$ positive roots, which says that the equation
$$nS_n-(n-1)binomn1S_n-1x+...+(-1)^n-1binomnn-1S_1x^n-1=0$$
has $n-1$ positive roots and by Rolle again the equation
$$left(nS_n-(n-1)binomn1S_n-1x+...+(-1)^n-1binomnn-1S_1x^n-1right)^(n-3)=0$$ or
$$S_1x^2-2S_2x+S_3=0$$
has two positive roots, which says that $S_2^2geq S_1S_3$.

Now we see that the equality occurs, when two roots of the equation $S_1x^2-2S_2x+S_3=0$

are equal, which says that all $a_i$ are equal.

If $a_i$ can be $0$ than the equality occurs in another cases and this thing we can check directly.