Prove that $P(1)P(xy) \ge P(x)P(y)$ where $P$ is a polynomial with positive real coefficients and $x\ge 1$ and $y\ge 1$

Question

I have one question:

Prove that $$P(1)P(xy) \ge P(x)P(y)$$

where $P$ is a polynomial with positive real coefficients and $x\ge 1$ and $y\ge 1$.

I already try to use the deviated of $P$ but it don't work.

I need help best regards

Answer 1

Consider function $f(x) = P(x)P(\frac{a^2}{x})$ where $a$ is a positive constant greater than 1. Then the derivative of $f'(x) = P'(x)P(\frac{a^2}{x}) - \frac{a^2}{x^2} P(x)P'(\frac{a^2}{x})$,

Then the second derivative of $f''(x) = P''(x)P(\frac{a^2}{x}) + \frac{a^4}{x^4} P(x)P''(\frac{a^2}{x})$

Since coefficients of polynomial $P$ are positive, the coefficients of $P'$ and $P''$ are also positive. Hence, $P(x)$ and $P''(x)$ is positive for $x \in [1,a]$.

From the above, it can be seen $f'(a) = 0$ and $f''(x) > 0$ for $x \in [1,a]$. Hence, $f(x)$ has a minima at $x = a$ and $f(x) >= f(a)$ for $x \in [1,a]$

Since the inequality is symmetric in $x$ and $y$, without the loss of generality, inequality can be expressed as $f(1) >= f(a)$ where $y = \frac{a^2}{x}$ which we have shown above.

Answer 2

Let $P(x)=a_0+\dots+a_nx^n$, then \begin{align} P(1)P(xy)&=\sum_{i<j}a_i a_j(x^jy^j+x^iy^i) + \sum_{i=0}^{n}a_i^2x^iy^i\\ P(x)P(y)&=\sum_{i<j}a_i a_j(x^iy^j+x^jy^i)+\sum_{i=0}^{n}a_i^2x^iy^i. \end{align} Notice that $x^jy^j+x^iy^i-x^iy^j-x^jy^i=(x^j-x^i)(y^j-y^i)\geq 0$ for $j>i$ and $x,y \geq 1$, so we have $$ P(1)P(xy)-P(x)P(y)=\sum_{i<j}a_i a_j(x^j-x^i)(y^j-y^i)\geq 0. $$ (the above is based on https://artofproblemsolving.com/community/c6h2391254p19626009)