Given: $$P(x)=x^4+a_0 x^3+a_1 x^2+a_2 x+a_3, P(x)>0, \forall x \in \mathbb R$$ $$\text{and}~~Q(x)=a_3x^4+a_2 x^3+a_1x^2+a_0 x$$ Prove or disprove: $Q(a_3)>-2$.
This is a question from a math contest. Working on the problem, using the assumptions, it is easy to see that $$Q(a_3)=a_3(a_3^4+a_2 a_3^2+a_1a_3+a_0)$$ and,
$$(a)~a_3>0,~~~(b)~1+a_0+a_1+a_2+a_3>0,~~~(c)~1-a_0+a_1-a_2+a_3>0$$ so that adding the inequalities (b) and (c) we find $$a_1+a_3>-1$$ I think that I'm close to untie the knot but I feel that I'm missing something.
Hints and solutions are appreciated.
If you observe carefully, you'll see that $${\rm P}(x)=x^4+a_0 x^3+a_1 x^2+a_2 x+a_3$$ $${\rm P}\left( \frac 1x \right)=\frac{a_3x^4+a_2 x^3+a_1x^2+a_0 x+1}{x^4}$$
$$\implies {\rm P}\left( \frac 1x \right)=\frac{{\rm Q}(x)+1}{x^4}$$
$${\rm Q}(x)=x^4 {\rm P}\left( \frac 1x \right) - 1$$
Now we have ${\rm P}(x) >0 \; \forall x \implies {\rm P}(0) >0 \implies \color{blue}{a_3 >0}$
and ${\rm P}\left(\dfrac 1{a_3}\right) >0$.
Therefore we have $$a_3^4 \; \cdot{\rm P}\left(\frac 1{a_3}\right) >0 \implies a_3^4 \; \cdot{\rm P}\left(\frac 1{a_3}\right)-1= \color{red}{{\rm Q}(a_3) >-1}$$
Thus ${\rm Q}(a_3)>-2$