Prove that $\frac{n-1}{n}>\frac{2a_0a_2}{a_1^2}$ 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
$$\frac{n-1}{n}>\frac{2a_0a_2}{a_1^2}$$
 A: 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.
A: Let $x_i, 1\le i\le n$ are the $n$ distinct real roots of the given equation.
Now, if $i\ne j,(x_i-x_j)^2>0\implies x_i^2+x_j^2>2x_ix_j$
There are $^nC_2$ such pairs of $(i,j)$, each $x_i$ occurs $n-1$ times in the LHS,
so $(n-1)(\sum x_i^2)>2\sum x_ix_j$
$(n-1)\{(\sum x_i)^2-2\sum x_ix_j\}>2\sum x_ix_j$
$\implies (n-1)(\sum x_i)^2>2n\sum x_ix_j$
Now, $\sum x_i=-\frac{a_1}{a_0}$ and $\sum x_ix_j=\frac{a_2}{a_0}$ using this or this.
