The answer for a general $n$ is positive: the discriminant is a sum of squares of polynomials in the entries of $H$. The first formula was given by Ilyushechkin and involves $n!$ squares. This number was improved by Domokos into $$\binom{2n-1}{n-1}-\binom{2n-3}{n-1}.$$ See Exercise #113 on my page.

The answer for a general $n$ is positive: the discriminant is a sum of squares of polynomials in the entries of $H$. The first formula was given by Ilyushechkin and involves $n!$ squares. This number was improved by Domokos into $$\binom{2n-1}{n-1}-\binom{2n-3}{n-1}.$$ See Exercise #113 on my page.

Details of Ilyushechkin's solution. Consider the scalar product $\langle A,B\rangle={\rm Tr}(AB)$ over ${\bf Sym}_n({\mathbb R})$. It extends as a scalar product over the exterior algebra. Then the discriminant equals $$\|I_n\wedge H\wedge\cdots\wedge H^{n-1}\|^2,$$ which is a sum of squares of polynomials.