Find all positive integers $a, b, c$ and integers $x, y, z$ such that $$\begin{align*} ax^2+by^2+cz^2=abc+2xyz-1\\ ab+bc+ca\geq x^2+y^2+z^2 \end{align*}$$
My Progress - By $\pmod 4$. I could conclude that at least one of the $a, b, c$ has to be even to satisfy the parity of both sides.
In this equation:
$$aX^2+bY^2+cZ^2=abc+2XYZ+F$$
$F$ - integer number given by the condition of the problem. A rather Tran decision:
$$a=(2pk-p^2+p-k^2)((t-s)^2-1)+2tsk+p(1-t^2)-(2k-p+1)s^2+F$$
$$b=(2pk-p^2+p-k^2)((t-s)^2-1)+2tsk-pt^2-(2k-p+1)(s^2-1)+F$$
$$c=(t-s)^2-1$$
$$X=t$$
$$Y=s$$
$$Z=(2pk-p^2+p-k^2)((t-s)^2-1)+k(2ts+1)-pt^2-(2k-p+1)s^2+F$$
$p,s,k,t$ - integers asked us.