TITLE: Prove that $\frac{n-1}{n}>\frac{2a_0a_2}{a_1^2}$
QUESTION [6 upvotes]: Given that the following equation $$p(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n=0$$ has $n$ distinct real roots. Prove that
$$\frac{n-1}{n}>\frac{2a_0a_2}{a_1^2}$$

REPLY [2 votes]: Hint proceed by induction on $n$ beginning with the case $n = 2$, which basically reduces to the condition for when a quadratic polynomial has two distinct real roots.
Then if $f(x) = a_nx^n + \dots + a_0$ has n distinct real roots, the derivative
\begin{equation}
f'(x) = na_ox^{n-1} + (n-1)a_1x^{n-2} + (n-2)a_2x^{n-3} + \dots a_1
\end{equation}
will have $n-1$  distinct real roots by Rolle's theorem. Think about this: in each interval of consecutive roots of $f(x)$ there will be a point where the tangent line is horizontal, and there are $n-1$ intervals if you arrange the roots of $f(x)$ in increasing order. Apply induction hypothesis to $f'(x)$, and magically everything will pop out after some elementary algebra.
Let me know if this is too cryptic.