Inequalities in Polynomial Let $p(x)$ be a non constant polynomial with non zero real coefficients. let $p(x) = a_{n}x^{n} + \dots   + a_{0} $.If the roots of $p(x)$ are denoted by $x_{1} ,x_{2} \dots  , x_{n} $ and if all the roots are real , then prove that  $$\\
 \frac{a_{n-j}^{2}}{{n \choose j}a_{n}^{2}} \leq \frac{1}{j}(n-1) \mathop{\sum_{i=1}^{n}} x_{i}^{2j}$$  where $j\in N , 0 \leq j \leq n \geq 1 $
and $$\frac{1}{j}{n \choose j}^{{n \choose j}-1} (n-1) \mathop{\sum_{i=1}^{n}} x_{i}^{j{n \choose j} } \geq \Bigl| \frac{a_{n-j}^{j}}{a_{n}^{j}} \Bigl|$$
What is the equality case?  
I have no idea how to attack it.
 A: We will use Elementary symmetric polynomial
, to get the proof, we fix first some notation, so for $n\geq 1$ and $1\leq i\leq n$ we put :
$$
\sigma_i(x_1,x_2,\dots,x_n)= \sum_{1\leq j_1<j_2<\dots<j_i\leq n}x_{j_1}x_{j_2}\dots x_{j_i}
$$
and $(x_i)_{1\leq i \leq n}$ will  denote the zeros of $P(x)=\sum_{i=0}^n a_n x^i$, so Vieta's formulas give :
$$
\sigma_i(x_1,x_2,\dots,x_n)=(-1)^i \frac{a_{n-i}}{a_n}
$$
so 
\begin{eqnarray}
\left(\frac{a_{n-i}}{a_n}\right)^2= \left(\sigma_i(x_1,x_2,\dots,x_n)\right)^2&=& \left(\sum_{1\leq j_1<j_2<\dots<j_i\leq n}x_{j_1}x_{j_2}\dots x_{j_i}\right)^2\\
&=&\binom{n}{i}^2\left(\sum_{1\leq j_1<j_2<\dots<j_i\leq n}\frac{x_{j_1}x_{j_2}\dots x_{j_i}}{\binom{n}{i}}\right)^2\\
&\leq& \binom{n}{i}^2\sum_{1\leq j_1<j_2<\dots<j_i\leq n}\frac{x_{j_1}^2x_{j_2}^2\dots x_{j_i}^2}{\binom{n}{i}} \qquad \textrm{convexity of }x^2\\
&=& \binom{n}{i}\sum_{1\leq j_1<j_2<\dots<j_i\leq n}x_{j_1}^2x_{j_2}^2\dots x_{j_i}^2 \\
\end{eqnarray}
So If $i=1$ the inequality is obvious :
$$
\left(\frac{a_{n-1}}{a_n}\right)^2\leq \binom{n}{1}\sum_{1\leq j \leq n}x_{j}^2 \leq  \binom{n}{1}(n-1) \sum_{j=1}^n x_{j}^2 
$$
If now $j>1$ we need the following lemma, witch can be proved directly from  the Inequality of arithmetic and geometric means 

Let $(s_i)$ a sequence of non-negative numbers, let $n\in \mathbb{N}^*$ so :
  $$ n \prod_{i=1}^n s_i\leq \sum_{i=1}^n s_i^n$$

So 
\begin{eqnarray}
\left(\frac{a_{n-i}}{a_n}\right)^2&\leq& \binom{n}{i}\sum_{1\leq j_1<j_2<\dots<j_i\leq n}x_{j_1}^2x_{j_2}^2\dots x_{j_i}^2 \\
&\leq & \binom{n}{i}\frac{1}{i}\sum_{1\leq j_1<j_2<\dots<j_i\leq n} x_{j_1}^{2i}+x_{j_2}^{2i}+\dots +x_{j_i}^{2i}\\
&\leq& \binom{n}{i}\frac{n-1}{i}\sum_{j=1}^n x_{j}^{2i}\\
\end{eqnarray}
For the second formula using the same reasoning we can prove that :
$$
\left|\frac{a_{n-i}}{a_n}\right|^{\binom{n}{i}}\leq \binom{n}{i}^{\binom{n}{i}-1}\frac{n-1}{i}\sum_{j=1}^n x_{j}^{\binom{n}{i} i}
$$
The third question about equality case, if you see we have use the convexity of $x^2$ for the first inequality and $x^{\binom{n}{i}}$ for the second  but these function are strictly convex and then we have equality if and only if all roots of $P$ are equal (i.e) $P(x)=a_n (x-x_1)^n$. 
