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 $x=y$
 A: Even for $f(x)=x^n$ it's not so easy.
We need to prove 
$$(x+y)\Big(\frac{x^2+y^2}{x+y}\Big)^n(x^n+y^n)\geq 2\left(x^{n+1}+y^{n+1}\right)\Big(\frac{x+y}{2}\Big)^n$$ or
$$2^{n-1}(x^2+y^2)^n(x^n+y^n)\geq\left(x^{n+1}+y^{n+1}\right)(x+y)^{2n-1}.$$
Now, let $x=ty$. 
Also, since our inequality is symmetric, we can assume that $t\geq1.$
Thus, we need to prove that $g(t)\geq0,$ where
$$g(t)=(n-1)\ln2+n\ln(t^2+1)+\ln(t^n+1)-\ln\left(t^{n+1}+1\right)-(2n-1)\ln(t+1).$$
Now, $$g'(t)=\frac{h(t)}{(t^2+1)\left(t^{n+1}+1\right)\left(t^n+1\right)(t+1)},$$
where $$h(t)=n(t-1)^3(t+1)t^{n-1}+2n(t-1)\left(t^{2n+1}+1\right)-(t^2+1)\left(t^{2n}-1\right).$$
Now, prove that $$h(1)=h'(1)=h''(1)=0$$ and $h'''(t)\geq0$ for all $t\geq1.$
A: Here is a partial answer. I show below that your inequality holds when $n\leq 5$.
Let $d=(x+y)^n\frac{LHS-RHS}{(y-x)^4}$. This turns out to be a polynomial, and a computation in PARI-GP (see below) reveals that $d$ is a polynomial in positive coefficients in $x,y,a_0,a_1,\ldots,a_n$ when $n\leq 4$ ; I call $d$ a completely positive polynomial.
Things become more complicated for $n=5$, because we have a "negative component" equal to
$$
-\frac{35}{16}a_0a_5(x^4y^3+x^3y^4)
$$
However, we are still able to obtain positivity of $d$ in this case because the $a_0a_5$ coefficient is of the form
$(y-x)^2$ times something completely positive.
As $n$ grows, the negative monomials grow in number too but they remain a minority, so the method for $n=5$ can probably be generalized.
Here is the PARI-GP code I used :
n0=4
aa(k)=eval(Str("a",k))
expr1=sum(j=0,n0,aa(j)*(u^j))
expr2=subst(expr1,u,p/q)*(q^n0)
expr3=subst(subst(expr2,p,x^2+y^2),q,x+y)
exprx=subst(expr1,u,x)
expry=subst(expr1,u,y)
main0=(x+y)*expr3*(exprx+expry)-\
2*((x+y)^n0)*(x*exprx+y*expry)*(subst(expr1,u,(x+y)/2))
main=main0/((y-x)^4)

A: I have tried many approaches to prove this inequality, but none worked. By now, I think that the inequality doesn't hold and thus I started to search for a counterexample. Michael Rozenberg has gave a proof for the special case $f(x) = x^n$ and Ewan Delanoy verifies this inequality for polynoms with degree at most $5$.

First we note that the condition that all $a_i >0$ are positive is unnecessary: If the inequality is valid for all $a_i> 0$, then we could let $a_i \downarrow 0$ for any index $i =0,\ldots,n$ and the inequality would remain to be valid.

However, it is in general wrong: Let $x=1$ and $y=t$ and take $f(x) = 1+x^{10}$. Then the function
$$g(t):= (1+t) f \Big( \frac{t^2+1}{1+t} \Big)(f(1)+f(t)) - 2 (f(1)+t f(t)) f \Big( \frac{t+1}{2} \Big) $$
is negative for $t=0.5$: See this plot in WolframAlpha.

Conjecture: Is the inequality valid if we additionally require that $a_0 \ge a_1 \ge ... \ge a_n$?

I couldn't find any counterexample on this strengthened variant, but also  no promising approach to prove this. Maybe someone has an idea?
A: A partial solution
Let $u=\frac{x}{x+y},v=\frac{y}{x+y}$. Then the inequality becomes $$f(ux+vy)(\frac{1}{2}f(x)+\frac{1}{2}f(y))\geq (uf(x)+vf(y))f\Big(\frac{x+y}{2}\Big)$$
In this form it is clear that the inequality depends upon the relative "importance" of two applications of Jensen's inequality.
A: Conjecture:
$$(x+y)f_n\Big(\frac{x^2+y^2}{x+y}\Big)(f_n(x)+f_n(y))- 2(xf_n(x)+yf_n(y))f_n\Big(\frac{x+y}{2}\Big) = \frac{(x-y)^4}{(x+y)^{n-1}}g_m(x,y)$$
with $g_m(x,y) \ge 0$ a polynomial. 
A: 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 $f(x)=x^n$. 
A: Edit: the proof is wrong in the last line, there is a counterexample to the statement in another answer.
I think I have a complete and easy to follow proof.
First let's define some quantities: $S_k(x, y) = x^k + y^k$, $S^f_k(x, y) = S_k(f(x), f(y))$ and $T_f(x, y) = xf(x) + yf(y)$.
Your inequality is trivial equivalent to:
$$f\left(\frac{S_2}{S_1}\right) \frac{S_1^f}{S_0} \ge f\left(\frac{S_1}{S_0}\right) \frac{T_f}{S_1}$$
Now $f = \sum_i a_i x^i$: expanding the expression of $f$ in both sides, we are left with the inequality:
$$\sum_{i,j} a_i \left(\frac{S_2}{S_1}\right)^i a_j \frac{S_j}{S_0} \ge \sum_{i,j} a_i \left(\frac{S_1}{S_0}\right)^i a_j \frac{S_{j+1}}{S_1}$$
Since $a_i, a_j > 0$ it is enough to prove that $\left(\frac{S_2}{S_1}\right)^i \frac{S_j}{S_0} \ge \left(\frac{S_1}{S_0}\right)^i \frac{S_{j+1}}{S_1}$.
We will prove that
$$\left(\frac{S_1^2}{S_2 S_0}\right)^i \le 1 \le \frac{S_1 S_j}{S_0 S_{j+1}}$$
For the LHS we have $S_1^2 = (x + y)^2 \le 2 (x^2 + y^2) = S_0 S_2 \Leftrightarrow 2xy \le x^2 + y^2$ which is true because of the square inequality, thus $\text{LHS} \le 1$.
For the RHS we must prove that $(x + y)(x^j + y^j) \ge 2 (x^{j+1} + y^{j+1})$ which as before is true if and only if $x^j y + y^j x \ge x^{j+1} + y^{j+1}$ and this is equivalent to the factorization $(x-y)(x^j - y^j) \le 0$ which is true for all choices of $x, y$ since the power operation preserves ordering.
