Exercise 18. Let $d \in \mathbb{N}^{*}$. Let $\mathrm{I}$ be an interval of the form $] \mathrm{a}, \mathrm{b}[$ with $\mathrm{a}<\mathrm{b}$. We define the length of $\mathrm{I}$ as $b-a$. We say that $\mathrm{I}$ is fantastic if for every polynomial $P$ of the form $P(x) = x^{2 \mathrm{~d}} + \sum_{i=0}^{2 \mathrm{~d}-1} a_{i} x^{\mathrm{i}}$ with $\mathrm{a}_{0}, \ldots, \mathrm{a}_{2 \mathrm{~d}-1}$ in $\mathrm{I}$, $P$ has no real roots.

Determine the maximum length of a fantastic interval, and determine all fantastic intervals of maximum length.
Solution to Exercise 18 Let's note that the condition is somewhat artificial: not having real roots is not particularly practical. Therefore, we use the following lemma:

Lemma 1: Let $\mathrm{P}$ be a polynomial with leading coefficient 1. $\mathrm{P}$ does not have real roots if and only if $P(x) > 0$ for all real $x$.

Proof. Indeed, let's reason by contraposition. If $\mathrm{P}$ has a real root, we have $\mathrm{P}(\mathrm{r}) = 0$, so $P(r) \leqslant 0$ for some real $r$. If $P(r) \leqslant 0$ for some real $r$, by its leading coefficient, $P$ tends to $+\infty$ as $x \to +\infty$. By the continuity of $\mathrm{P}$ and the intermediate value theorem, $\mathrm{P}$ has a root in $[\mathrm{r}, +\infty[$.

Thus, this simplifies the condition in the statement: any polynomial formed with the constraints of the statement is strictly positive on $\mathbb{R}$.

Lemma 2: Let $I$ be a fantastic interval and $a < b$ be two real numbers in $I$. Then $P(x) > 0$ for all $x \in \mathbb{R}$ according to the first lemma. Note that the inequality is true if $x \geqslant 0$.

If $x < 0$, we need to show that:

$$
d|x|^{2 d} + d \sum_{i=0}^{d-1}|x|^{2 i} - (d+1) \sum_{i=0}^{d-1}|x|^{2 i+1} > 0
$$

By the arithmetic-geometric mean inequality, $\frac{|x|^{2 i} + |x|^{2(i+1)}}{2} \geqslant |x|^{2 i+1}$ for $0 \leqslant i \leqslant d-1$.

By summing, we obtain:

$$
\frac{d}{2}\left(1 + |x|^{2 d}\right) + d \sum_{i=1}^{d-1}|x|^{2 i} = \frac{d}{2} \sum_{i=0}^{d-1}|x|^{2 i} + |x|^{2(i+1)} \geqslant d \sum_{i=0}^{d-1}|x|^{2 i+1}
$$

To show that $\mathrm{dP}(\mathrm{x}) > 0$, it suffices to show the following inequality:

$$
\frac{d}{2}\left(1 + |x|^{2 d}\right) \geqslant \sum_{i=0}^{d-1}|x|^{2 i+1}
$$

By the weighted arithmetic-geometric mean inequality, for all $i$ between 0 and $d-1$, $\frac{2 d - (2 i + 1)}{2 d} \times 1 + \frac{2 i + 1}{2 d}|x|^{2 d} \geqslant |x|^{2 i+1}$.

By summing, since the sum of the odd numbers between 1 and $2 d - 1$ is $d^2$, we obtain:

$$
\frac{d^2}{2 d} \times 1 + \frac{d^2}{2 d}|x|^{2 d} = \frac{d}{2}(1 + |x|)^{2 d} \geqslant \sum_{i=0}^{d-1}|x|^{2 i+1}
$$

which gives the desired result.

Thus, $] 1, 1 + \frac{1}{d} [$ is the unique fantastic interval of maximum size.