An inequality for polynomials with positives coefficients I have found in my old paper this theorem :

Let $a_i>0$ be real numbers and $x,y>0$ then we have :
  $$(x+y)f\Big(\frac{x^2+y^2}{x+y}\Big)(f(x)+f(y))\geq 2(xf(x)+yf(y))f\Big(\frac{x+y}{2}\Big)$$
  Where :$$f(x)=\sum_{i=0}^{n}a_ix^i$$

The problem is I can't find the proof I made before . Furthermore I don't know if it's true but I have checked this inequality a week with Pari\Gp and random polynomials defined as before . 
So first I just want a counter-examples if it exists .
If it's true if think it's a little bit hard to prove . I have tried the power series but without success . 
Finally  it's a refinement of Jensen's inequality for polynomials with positives coefficients .
Thanks a lot if you have a hint or a counter-example .
Ps:I continue to check this  and the equality case is to $x=y$

First we rewrite your inequality as
$$ \frac{\frac{1}{2}f(x) +\frac{1}{2}f(y)}{f\left(\frac{1}{2}x + \frac{1}{2}y\right)} \ge \frac{\frac{x}{x+y}f(x) +\frac{y}{x+y}f(y)}{f\left(\frac{x}{x+y}x + \frac{y}{x+y}y\right)}. $$
For $\lambda \in [0, 1]$, let $$ g(\lambda) = \frac{\lambda f(x) + (1 - \lambda) f(y)}{f(\lambda x + (1 - \lambda)y)}. $$
Since $f(x)$ is convex, we know $g(\lambda) \ge 1$. It suffice to show $g\left(\frac{x}{x+y}\right) \le g\left(\frac{1}{2}\right)$. Without losing generality, let's assume $x > y$. Furthermore, it suffice to show $g(\lambda)$ is a non-increasing function for $\lambda \in [\frac{1}{2}, \frac{x}{x+y}]$.
$$g^\prime(\lambda) = \frac{f(x) - f(y)}{f(\lambda x + (1 - \lambda)y)} - \frac{\lambda f(x) + (1 - \lambda) f(y)}{{\left(f(\lambda x + (1 - \lambda)y)\right)}^2}f^\prime(\lambda x + (1 - \lambda)y)(x-y).$$
For $g^\prime(\lambda)$, we are only concerned about its being positive or not. To simplify notation, we use $\stackrel{s}{=}$ to denote sign equality so that we can drop nonnegative terms.
We first consider the special case of $f(x)=x^n, n \ge 1$ (the case $n=0$ is trivial), then $ \frac{f^\prime(x)}{f(x)}=\frac{n}{x}.$ Hence
$$ g^\prime(\lambda) = \frac{1}{{(\lambda x + (1 - \lambda)y)}^n}\left( 
{x^n - y^n} - n(x-y)\frac{\lambda x^n + (1-\lambda) y ^n}{\lambda x + (1 - \lambda)y} \right) \\ \stackrel{s}{=} (x^n - y^n)(\lambda x + (1 - \lambda)y) - n(x-y)\left(\lambda x^n + (1-\lambda) y ^n\right) \\ = (x^n - y^n)\left(\frac{x+y}{2}\right) - n(x-y)\left(\frac{x^n + y ^n}{2}\right)   \quad\quad\text{(1)} \\ - (n-1)\left(\lambda - \frac{1}{2}\right)(x-y)(x^n - y ^n) \quad\quad\text{(2)}.$$
For $\lambda \in [\frac{1}{2}, \frac{x}{x+y}]$, it is easy to see (2) is non-positive. Note (1) is non-positive as well since
$$ (x^n - y^n)\left(\frac{x+y}{2}\right) - n(x-y)\left(\frac{x^n + y ^n}{2}\right)  \le 0 \\ \Longleftrightarrow (x^n - y^n)\left({x+y}\right) \le n(x-y)\left({x^n + y ^n}\right) \\ \Longleftrightarrow x^{n+1} - y^{n+1} -xy^n +yx^n \le n(x^{n+1}-y^{n+1}+xy^n-yx^n) \\ \Longleftrightarrow (n+1)\left(x^ny-xy^n\right) \le (n-1)\left(x^{n+1}-y^{n+1}\right) \\ \Longleftrightarrow (n+1)\left(z^n-z\right) \le (n-1)\left(z^{n+1}-1\right), z=\frac{x}{y}>1 \\ \Longleftrightarrow \ell(z) =  (n-1)\left(z^{n+1}-1\right) - (n+1)\left(z^n-z\right) \ge 0. \text{ for } z>1 $$
The last statement is true since $\ell(1) = 0$ and 
$$\ell^\prime(z) = (n^2-1)z^n-(n+1)(nz^{n-1}-1)\\ = (n+1)\left(n(z^n-z^{n-1}) - (z^n - 1)\right)\\ = (n+1)(z-1)\left( nz^{n-1}-\sum_{i=0}^{n-1}z^i \right) \ge 0 \text{ for } z \ge 1.$$
For the case of $f(x)=x^n$, we have proven the inequality. I will try to generalize the conclusion to all polynomials with positive coefficients later when I get time.

Noticed that @p4sch put in a counter example for general polynomial cases, I will hence abandon any generalization efforts, and my proof is only valid for the special case of $f(x)=x^n$. 
