So consider both the equations $x^2+ax+b=0$ and $x^2+bx+a=0$, has one root, that is not distinct roots Then we have $a^2=4b$ and $b^2=4a$, considering the discriminants to be zero Then multiplying both, we get $(ab)^2=16(ab)$, and $(ab)(ab-16)=0$, so $ab=16$, but we have $a>4, b>4$, so $ab>16$, contradiction!! Then at least one of the two trinomials must have distinct roots (Proved)

$\Delta_1= a^2-4b,\Delta_2 =b^2-4a$ $\Delta_1+\Delta_2=a^2+b^2-4a-4b\ge \frac12 (a+b)^2-4(a+b)=(a+b)(\frac 12(a+b)-4) $ but we know$\frac12(a+b)>4$ then $\Delta_1+\Delta_2>0$ hence one of them is greater than $0$