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it's a conjecture on polynomials with positive coefficient :

Let $x,y>0$ then we have : $$\frac{x^2f(x)+y^2f(y)}{xf(x)+yf(y)}\leq \frac{x^2+y^2}{x+y}\frac{f\Big(\frac{x^3+y^3}{x^2+y^2}\Big)}{f\Big(\frac{x^2+y^2}{x+y}\Big)}$$ Where ($k$ a natural number): $$f(x)=\sum_{k=0}^{n}a_kx^k$$ And :$$a_k>0 \quad\forall k\geq0$$

The problem of my previous inequality An inequality for polynomials with positives coefficients was the fixed coefficient $2$ . To recall I just apply Jensen's inequality on the denominator and the numerator of the LHS .

Furthermore I have tried exotic polynomials like :

$$f(x)=\sin(1)+ex^3+x^{10}$$

I have tried also monomials .

Polynomials like :

$$f(x)=1+\frac{1}{2}x+\frac{1}{3}x^2+\cdots +\frac{1}{n}x^{n-1}$$

Works also .

And the counter-example $f(x)=1+x^{10}$ works .

So my question : Have you a counter-example ?

Thanks a lot for sharing your time and knowledge .

Ps:I work with Pari-Gp.

Barackouda
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1 Answers1

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$$\text{Counterexample}$$$$\boxed{x=0.5,\quad y=1.01,\quad f(y):=1+0.0001(y+y^2+\cdots+y^{99})+y^{100}}$$ $$\frac{0.5^2f\left(0.5\right)+1.01^{2}f\left(1.01\right)}{0.5f\left(0.5\right)+1.01f\left(1.01\right)}>0.95,\quad\frac{0.5^2+1.01^{2}}{0.5+1.01}\cdot\frac{f\left(\frac{0.5^{3}+1.01^{3}}{0.5^2+1.01^{2}}\right)}{f\left(\frac{0.5^2+1.01^{2}}{0.5+1.01}\right)}<0.85$$

TheSimpliFire
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