I know that $ \forall n\in N, a_0 + a_1 + ... + a_n = 1$, with $a_0, a_1, ... a_n > 0$ and $f(t) = a_0t^n+a_1t^{n-1}+...+a_n, \forall t\in R$
I have to prove that for every $x > 0$ $$ f^2\left(\sqrt{x^3} \right)* f\left(\frac1{x^3} \right)\ge1 $$
I thought that 1 from that relationship is $a_0+a_1+...+a_n$ And I also thought at Holder Inequality, but I do not know for sure if it's good.
Holder says : $$ \sum_{i=0}^n \Delta_i\alpha_i\beta_i \le \left( \sum_{i=0}^n \Delta_i\alpha_i^p \right)^{\frac1p}*\left(\sum_{i=0}^n \Delta_i\beta_i^q\right)^{\frac1q} $$
where p,q > 0, $\frac1p+\frac1q=1$ and $\Delta_i, \alpha_i, \beta_i \in R_+ \forall 0\le i \le n, \forall n\in N$
I had to proceed like this:
I noted $x^3 = t$ and so $\sqrt{x^3} = \sqrt t = t^{\frac12}$ and $\frac{1}{x^3}=\frac1t = t^{-1}$
I said p = $\frac12$, because $f\left(\sqrt{t}\right)=\left(a_0t^{pn} + a_1t^{p(n-1)}+...+a_{n-1}t^{p}+a_n \right)^{\frac1p}$ with $\Delta_i = a_i, \alpha_i=t^{n-i}$
I also believe that q should be -1 to verify $\frac1p+\frac1q=1$, but here it is 1, because $f\left(\frac1t \right) = \left(a_0t^{-n} + a_1t^{-(n-1)}+...+a_{n-1}t^{-1}+a_n \right)$ with $\Delta_i = a_i, \beta_i=t^{-(n-i)}$
- If I would get it right, I should get to $\sum_{i=0}^n \Delta_i\alpha_i\beta_i = a_0 + a_1 + ... + a_n = 1$ and the inequality would be proven.
So, is it correct what I am doing? How can I make q = -1? Or what am I getting wrong and makes me get stuck?
$f^2(x^{3/2})f(x^{-3})=(\sum a_{n-i}t^{3i/2})^2(\sum a_{n-i}t^{-3i})\ge \{\sum (a_{n-i}^3t^{3i/2}t^{3i/2}t^{-3i})^{1/3}\}^{3}=1$
All you need is :
($\sum x_i^2 y_i)^3\le(\sum x_i^3)^2(\sum y_i)$ (which is Holder inequality)
where $x_i=a_{n-i}t^{3i/2},y_i=a_{n-i}t^{-3i}$