Newton's Inequality's Equality case 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? 
 A: 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
$$x^n-\binom{n}{1}S_1x^{n-1}+...+(-1)^n\binom{n}{n}S_n=0.$$
Thus, the equation
$$S_nx^n-\binom{n}{1}S_{n-1}x^{n-1}+...+\binom{n}{n}(-1)^n=0$$ has $n$ positive roots and by Rolle the equation
$$\left(S_nx^n-\binom{n}{1}S_{n-1}x^{n-1}+...+\binom{n}{n}(-1)^n\right)'=0$$ or
$$nS_nx^{n-1}-(n-1)\binom{n}{1}S_{n-1}x^{n-2}+...+(-1)^{n-1}\binom{n}{n-1}S_1=0$$
has $n-1$ positive roots, which says that the equation
$$nS_n-(n-1)\binom{n}{1}S_{n-1}x+...+(-1)^{n-1}\binom{n}{n-1}S_1x^{n-1}=0$$
has $n-1$ positive roots and by Rolle again the equation
$$\left(nS_n-(n-1)\binom{n}{1}S_{n-1}x+...+(-1)^{n-1}\binom{n}{n-1}S_1x^{n-1}\right)^{(n-3)}=0$$ or
$$S_1x^2-2S_2x+S_3=0$$
has two positive roots, which says that $S_2^2\geq S_1S_3$.
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.
