# inequality – Proving on the equation \$(x^2+mx+n)(x^2+px+q)=0\$

Find all real numbers k such that if $$a,b,c,d in mathbb R$$ and $$a>b>c>d geq k$$ then there exist permutations $$( m ,n ,p,q)$$ of $$(a,b,c,d)$$ so that the following equation has 4 distinct real solutions :

$$(x^2+mx+n)(x^2+px+q)=0$$

Here all i did :

$$(x^2+mx+n)(x^2+px+q)=0$$

$$Leftrightarrow x^2+mx+n=0$$ or $$x^2+px+q=0$$

so I think the four real solutions, if any, of the equation can only be :

• $$x= frac{sqrt{m^2-4n} -m}{2}$$

• $$x= frac{-sqrt{m^2-4n} -m}{2}$$

• $$x= frac{sqrt{p^2-4q} -p}{2}$$

• $$x= frac{-sqrt{p^2-4q} -p}{2}$$

So I think to solve the problem we just need to find all real numbers k such that if $$a,b,c,d in mathbb R$$ and $$a>b>c>d geq k$$ then there exist permutations $$( m ,n ,p,q)$$ of $$(a,b,c,d)$$ so that
$$p^2 geq 4q$$ and $$m^2 geq 4n$$

We can prove that for $$k >= 4$$ it is absolutely true. So are there any other satisfying $$k$$ values ? I’m not entirely sure. Hope to get help from everyone. Thanks very much !