Universal sums of three quadratic polynomials

Let $a,b,c,d,e$ and $f$ be integers with $a\ge c\ge e>0$, $b>-a$ and $b\equiv a\pmod2$,$d>-c$and$d\equiv c\pmod 2$,$f>-e$and$f\equiv e\pmod2$. Suppose that $b\ge d$ if $a=c$, and $d\ge f$ if $c=e$. When $b(a-b)$, $d(c-d)$ and $f(e-f)$ are not all zero, we prove that if each $n\in\mathbb N={0,1,2,\ldots}$canbewritten$x(ax+b)/2+y(cy+d)/2+z(ez+f)/2$with$x,y,z\in\mathbb N$thenthetuple$(a,b,c,d,e,f)$mustbeonourlistof$473$candidates,andshowthat56ofthemmeetourpurpose\mathrm{.}When$b\in[0,a)$,$d\in[0,c)$and$f\in[0,e)$,weinvestigatetheuniversaltuples$(a,b,c,d,e,f)$over$\mathbb Z$forwhichany$n\in\mathbb N$canbewritten$x(ax+b)/2+y(cy+d)/2+z(ez+f)/2$with$x,y,z\in\mathbb Z$, and show that there are totally 12082 such candidates some of which are proved to be universal tuples over $\mathbb{Z}$. For example, we show that any $n\in \mathbb{N}$ can be written as $x(x+1)\mathrm{/}2+y(3y+1)\mathrm{/}2+z(5z+1)\mathrm{/}2$ with $x,y,z\in \mathbb{Z}$, and conjecture that each $n\in \mathbb{N}$ can be written as $x(x+1)\mathrm{/}2+y(3y+1)\mathrm{/}2+z(5z+1)\mathrm{/}2$ with $x,y,z\in \mathbb{N}$.