If $\ a_1,\ a_2,\ a_3,...\ a_n$ $n\ge2$are real and $(n-1)\ a_1^2 -2n\ a_2 <0$ prove that at least two roots of the equation $x^n+\ a_1x^{n-1}+\ a_2x^{n-2}+...+\ a_n=0$ are imaginary.
Manually i am able to prove it as $\ a_2$ is always positive. I tried different combination and it worked but needs logical proof from the equation.
Suppose first that all roots of the polynomial $f=x^n+ a_1x^{n-1}+ a_2x^{n-2}+\dots+ a_n$ are real and distinct. Then the derivative $f'$ has also all roots real and distinct, they are placed in the open intervals between consecutive roots. Inductively, we build the derivative till we get a polynomial of degree two. This polynomial is $$ g= f^{(n-2)}(x) = \frac {n!}{2!} \cdot x^2 + \frac {(n-1)!}{1!}\cdot a_1x + \frac {(n-2)!}{0!}\cdot a_2 \ . $$ We compute $g/(n-2)!$, which is $\frac 12n(n-1)x^2+(n-1)a_1x+a_2$, it has discriminant $<0$ by the given relation, so no real roots. Contradiction.
Our asumption is false.
The case when there are multiple roots is covered analogously, either by considering a deformation of the roots, or by observing that if $f$ has a root with algebraic multiplicity $r$, then the root survives with multiplicity $(r-1)$ in the derivative.