Inequalities involving polynomials Let $A_k=\{x \mid  2^k < x^k +x^{k+1} < 2^{k+1}\},\, k=1, 2, 3, \ldots$.
Find the largest positive integer $n$ such that 
$$
A_1\cap A_2\cap\cdots \cap A_n \ne \emptyset.
$$
 A: $A_1\cap A_2\cap A_3\neq \emptyset$, but $A_1\cap A_4=\emptyset$.  I have no analytic proof, just a calculation on wolframalpha:
$A_4\approx(1.58,1.83)$, while $A_1\approx (-2.56,-2)\cup (1,1.56)$
Also $A_2\approx (1.31, 1.72)$ and $A_3\approx(-2.31,-2)\cup(1.48,1.79)$, so $A_1\cap A_2\cap A_3\approx(1.48,1.56)$
Proof that $A_1\cap A_2\cap A_3\neq \emptyset$:  Take $x=1.5$. $1.5+1.5^2=3.75$ and $2<3.75<4$.  $1.5^2+1.5^3=5.625$ and $4<5.625<8$.  $1.5^3+1.5^4=8.4375$ and $8<8.4375<16$.
Proof that $A_1\cap A_4=\emptyset$: Take $x=1.57$. 
 On the interval $(0,+\infty)$, $f(x)=x+x^2$ is increasing and $f(1.57)=4.0349>4$.  But also $g(x)=x^4+x^5$ is increasing on the same interval, and $g(1.57)=15.6146312657<16$.  Hence $1.57$ is strictly above $A_1$ and strictly below $A_4$.
A: Computers these days can do exact arithmetic, so there's no reason you can't have a computer do an "analytic" proof - or at least have it do the relevant computations.
The positive portions of your sets
\begin{array}{l}
 2<x^2+x<4 \\
 4<x^3+x^2<8 \\
 8<x^4+x^3<16 \\
 16<x^5+x^4<32 \\
\end{array}
look something like so

To show that the intersection of the first and last is empty, it suffices to find $x_0$ such that $x_0^2+x_0>4$ (i.e., $x_0$ lies to the right of the first interval) and $x_0^5+x_0^4<16$ (i.e., $x_0$ lies to the left of the last interval).  I propose that you choose
$$x_0 = \frac{196377467}{125086164}.$$
Then
$$x_0^2+x_0 - 4 = \frac{542019891463093}{15646548424234896} > 0$$
and
$$x_0^5+x_0^4 -16= -\frac{11888789705707430179748265315192519808033}{30622903893638169223375610368116647629824} < 0.$$
I obtained the number $x_0$ by simply rationalizing the average of numerical approximations to the two relevant endpoints, but the exact computation with rational numbers proves that $x_0$ has the desired properties.
