I'm sorry, but I have trouble: $Q(x)=ax^t+bx^s+cx^r$ I think, but then $Q(x)= x^3-x^2+x$ satisfy first observations, but $P(2.1)+P(-2.09)+P(-0.009)$ gives sth wrong, isn't it? ** Is the defenition of $Q(x)$ wrong and can you explain the question?

kroki forgot, from his initial post http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=480621&hilit=trinomial, where he used the expression "square trinomial", that it was meant for $Q$ to be a quadratic trinomial, thus $Q(x) = ax^2+bx+c$.

By the increasing property of $P(x)$ we know $a,c >0$ and $|32ac| \ge |9b^2|$, because $P(0)=0$ when $x,y,z>0$ or $b\ge 0$, we are done trivially. Otherwise, when two are negative, we have to prove $P(x+y)+(-y)+P(-x)\ge 0$ where $x,y>0.$ When we work this out, it follows by AM-GM $a(x+y)^4+c(x+y)^2\ge 0.75b(x+y)^3\ge 3bxy(x+y)$ and a few positive parts $a(x^4+y^4)+c(c^2+y^2)$.

It is required that \[ 4ax^3+3bx^2+2cx >0 \hspace{3mm} \forall x \in (0,\infty) \] or \[ 4ax^2+3bx+2x >0 \hspace{3mm} \forall x \in (0,\infty) \] Therefore, $a >0$ and $ \Delta=9b^2-32ac <0$. This also infers that $c >0$. Back to our expression \[ \sum_{cyc}x^2P(x)=a\sum_{cyc}x^4+b\sum_{cyc}x^3+c\sum_{cyc}x^2 \ge \sum_{cyc}\left(2\sqrt{ac}|x|^3+bx^3\right)>\sum_{cyc}\left(|bx^3|+bx^3\right)\ge 0\] Hence, we are done. P/s. Hope there is no misinterpretation here.

great math wrote: $9b^2-32ac <0$ $ 2\sqrt{ac}|x|^3+bx^3 >|bx^3|+bx^3$ This would mean the question is trivial and $P(x)\ge 0$ holds always. byt in fact yours gives $36ac > 9b^2$ instead of $32ac.$

$4ax^2+3bx+2c>0$ holds for every $x>0$, not for every $x\in R$,so $\Delta<0$ is incorrect. But I haven't an idea yet.

see the attachment

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great math wrote: It is required that \[ 4ax^3+3bx^2+2cx >0 \hspace{3mm} \forall x \in (0,\infty) \] or \[ 4ax^2+3bx+2x >0 \hspace{3mm} \forall x \in (0,\infty) \] Therefore, $a >0$ and $ \Delta=9b^2-32ac <0$. This also infers that $c >0$. Back to our expression \[ \sum_{cyc}x^2P(x)=a\sum_{cyc}x^4+b\sum_{cyc}x^3+c\sum_{cyc}x^2 \ge \sum_{cyc}\left(2\sqrt{ac}|x|^3+bx^3\right)>\sum_{cyc}\left(|bx^3|+bx^3\right)\ge 0\] Hence, we are done. P/s. Hope there is no misinterpretation here. How did you get this? $4ax^3+3bx^2+2cx >0 \hspace{3mm} \forall x \in (0,\infty)$

Here's a different approach that uses basic properties of an increasing function. If $x, y, z>0\leadsto f(x), f(y), f(z)>f(0)=0$ and we're done. So let $x, y<0$ and $z>0$. And since $z>-(x+y)>0\leadsto f(z)>f(-x-y)$. It's enough to show \begin{align*}f(x)+f(y)+f(-x-y)>0\hspace{4 mm} \textsf{for $x, y<0$} \end{align*}Or \begin{align*} f(-x)+f(-y)+f(x+y)>0\hspace{4 mm} \textsf{for $x, y>0$} \end{align*}Since $f(x)=ax^{4}+bx^{3}+cx^{2}$ is increasing $\leadsto a>0$. Also \begin{align*} f'(x)=4ax^{3}+3bx^{2}+2cx>0 \hspace{2 mm} \textsf{so} \hspace{2 mm} -2c<4ax^{2}+3bx \hspace{2 mm} \textsf{and} \hspace{2 mm} c\ge 0 \end{align*}We have $$f(-x)+f(-y)+f(x+y)=2a(x^{4}+y^{4}-x^{2}y^{2})+2c(x^{2}+y^{2})+xy(4a(x+y)^2+3b(x+y)+2c) \hspace{3 mm} (\star)$$And surprisingly $f'(x+y)=4a(x+y)^3+3b(x+y)^2+2c(x+y)>0$ and hence each summand in $(\star)$ is positive. $\blacksquare$ edit : yes , $f(x)\equiv P(x)$