The discriminant of a quadratic polynomial is $$\Delta=\left(x_2 \left(x_3+x_5\right)+x_4 \left(x_1+x_5\right)+x_6\left(x_1+x_3\right) \right){}^2-4 \left(x_1 x_3+x_3 x_5+x_1 x_5\right) \left(x_2 x_4+x_4 x_6+x_2 x_6\right)$$ with $\{x_1,x_2,\ldots,x_6\} \in \mathbb{R}_{\ge0}$. Prove that also $\Delta \in \mathbb{R}_{\ge0}$ !
A proof was given below in an answer. The proof uses inequalities.
Further question
Now I want to know if a proof without inequalities is possible. If for example we could finally write $\Delta =a^2+b^2+c^2$ we could directly see that $\Delta \ge 0$. Any other form that avoids inequalities are also acceptable. Inequalities can be used as intermediate steps, but the final form should not contain them. Is a form possible without inequalities?
Denote $X=(x_1,x_3,x_5)$ and $Y=(x_2,x_4,x_6)$ and $S(X)=(x_1+x_3+x_5).$ Consider the bilinear form $$B(X,Y)=S(X)S(Y)-\langle X,Y\rangle.$$ Then$$\Delta=B(X,Y)^2-B(X,X)B(Y,Y)$$$$=(S(X)\|Y\|-S(Y)\|X\|)^2+(\|X\|\|Y\|-\langle X,Y\rangle)(2S(X)S(Y)-\langle X,Y\rangle-\|X\|\|Y\|)$$From Schwarz $\|X\|\|Y\|-\langle X,Y\rangle>0$ and therefore $$2S(X)S(Y)-\langle X,Y\rangle-\|X\|\|Y\|\geq 2(S(X)S(Y)-\|X\|\|Y\|)$$ Finally, the fact that $X$ and $Y$ have non negative coefficients implies that $S^2(X)\geq \|X\|^2$, or $S(X)\geq \|X\|$ as well as $S(Y)\geq \|Y\|.$ This proves $\Delta\geq 0.$
For $X,Y\neq 0$, equality $\Delta=0$ takes place if and only if when $\langle X,Y\rangle=\|X\|\|Y\|$ and $S(X)^2=\|X\|^2$ and $S(Y)^2=\|Y\|^2$, namely either $x_3=x_4=x_5=x_6=0$ or $x_1=x_2=x_5=x_6=0$ or $x_1=x_2=x_3=x_4=0.$
Edit. More generally, if $x$ and $y$ are vectors of $R^n$ with non negative components and with the natural Euclidean structure, defining $S(x)=x_1+\cdots+x_n$ then $B(x,y)=S(x)S(y)-\langle x,y\rangle$ still satisfies $B(x,y)^2\geq B(x,x)B(y,y)$. If you prefer matrices, introduce $I_n$ the identity matrix and $J_n$ the matrix whose all entries are $1$ and observe that $B(x,y)=x^T(J_n-I_n)y.$