Assume that $(a,b)$ is an answer of this problem, where $a<b$. Obviously, $a>0$, because if $a \leq 0$, then choose $a_0=a$ and $a_1,...,a_{2d-1}$ arbitrarily, then $P(0)=a \leq 0$ but $P(x)>0$ as $x$ goes to $+\infty $, so MVT implies that $P(x)$ has a (non-negative) real root, a contradiction. So, we have: $0<a<b$ . Then, for such $a,b$ define $Q(x)=x^{2d}+bx^{2d-1}+ax^{2d-2}+...+bx+a$, obviously, for all $x \in R$ we must have $Q(x) \ge 0$, because if for some $y$ we have $Q(y)<0$, then by tending $a_{2d-1},...,a_1$ to $b$ and tending $a_{2d-2},...,a_2,a_0$ to $a$, we conclude that $P(y)<0$, then again MVT gives a real root, contradiction. So, we conclude that $Q(-1) \ge 0$ which implies that $b \leq a+1/d$, now we prove that $b=a+1/d$ can occur, and we will find all such $a$. Note that the above polynomial is positive for all $x \ge 0$ so we need to prove that it is also positive for all negative $x$, so replacing $x$ by $-x$ and $R(x)$ by $Q(x)$, we need to prove that the below polynomial is non-negative for all positive $x$: $R(x)=x^{2d}-(a+1/d)x^{2d-1}+ax^{2d-2}-...-(a+1/d)x+a=(x-1)((dx^{2d-1}+(d-1)x^{2d-2}+(d-1)x^{2d-3}+...+2x^4+2x^3+x^2+x)/d - ax^{2d-2}-ax^{2d-4}-...-ax^2-a )=(x-1)H(x)$ Now, assume $a>1$, put $x=1+p$ where $p>0$ is a very small positive real number, then $x-1>0$ so we must have $H(x)\ge 0$, but $H(x)=d(1-a)+A-B$ where $A,B$ are expressions both multiple by $p$, so since $d(1-a)<0$ we can choose $p$ so small so that $H(x)<0$ and we reach a contradiction.If $a<1$, then choose $x=1-p$ where $p>0$ is a small positive real number. Then, $x-1<0$ so we must have $H(x) \leq 0$ , but pay attention that $H(x)=d(1-a)+C+D-E-F$ where $C,D,E,F$ are expressions all multiple by $p$, and since $d(1-a)>0$, so we can choose $p$ so small so that $H(x)>0$ and we reach a contradiction. So, if $b=a+1/d$ were to have a solution, we must necessarily have $a=1$, and this is true; $(1,1+1/d)$ is an answer, because in that case we need to prove that: $R(x)=x^{2d}-(1+1/d)x^{2d-1}+x^{2d-2}-...-(1+1/d)x+1 >0 $ which is equivalent to proving that: $(x^{2d}+x^{2d-2}+...+x^{2}+1)/(d+1) > (x^{2d-1}+...+x^3+x)/d $ Using repeatedly the inequality : $x^{2c} +1 \ge x^{2c-2d+1}+x^{2d-1} $, which is true for all natural numbers $c \ge d \ge 1$, the problem would be solved.

Thank you!!

Iran 3rd MO 2019/Algebra 3 wrote: We are given a natural number $d$. Find all open intervals of maximum length $I \subseteq \mathbb{R}$ such that for all real numbers $a_0,a_1,...,a_{2d-1}$ inside interval $I$, we have that the polynomial $P(x)=x^{2d}+a_{2d-1}x^{2d-1}+...+a_1x+a_0$ has no real roots. Another amazing problem. We claim that the only open interval that satisfies the requirement is the interval $\left( 1, 1 + \frac{1}{d} \right)$. Claim 01. For all real numbers $r$, then $P(r) > 0$. Proof. We are immediately done if there exists any real number $r$ such that $P(r) = 0$. Now, if there exists real number $r$ such that $P(r) < 0$, note that $\lim_{x \to \infty} P(x) = \infty$ as $\text{deg} \ P$ is even. Now, by Intermediate Value Theorem, there must exists a real number $r_0 \in (r,\infty)$ such that $P(r_0) = 0$, a contradiction. Claim 02. The interval of maximum length has length not greater than $\frac{1}{d}$. Proof. Suppose that the open interval $I = (k, \ell)$ for some $k < \ell$. Consider $P(-1)$, where $a_{\text{odd}} = \ell - \varepsilon$ and $a_{\text{even}} = k + \varepsilon$. Then, \[0 < P(-1) = 1 + d (k + \varepsilon - \ell + \varepsilon) \]for all $\varepsilon > 0$. Therefore, taking $\varepsilon \to 0$, we conclude that $d(k - \ell) \ge -1$, which forces $\ell - k \le \frac{1}{d}$. Claim 03. There are no other open interval of length $\frac{1}{d}$ other than $I = \left( 1, 1 + \frac{1}{d} \right)$ that works. Proof. Let the open interval be $\left( \Delta, \Delta + \frac{1}{d} \right)$. We easily obtain $\Delta > 0$, by taking $P(0)$ and $a_0 = \Delta + \varepsilon$, where $\varepsilon \to 0$. Now, consider the polynomial \[ P(x) = x^{2d} + (\Delta + \varepsilon) \sum_{j = 0}^{d - 1} x^{2j} - \left( \Delta + \frac{1}{d} - \varepsilon \right) \sum_{i = 1}^{d} x^{2i - 1} \]We need to find $\Delta$ such that $P(x) > 0$ for all $x \in \mathbb{R}$. By taking $\varepsilon \to 0$, it suffices to have \[ Q(x) = x^{2d} + \Delta \sum_{j = 0}^{d - 1} x^{2j} - \left( \Delta + \frac{1}{d} \right) \sum_{i = 1}^d x^{2i - 1} \ge 0 \]for all $x \in \mathbb{R}$. Notice that $Q(1) = 0$. Since $Q(x) \ge 0$ for all $x \in \mathbb{R}$, then $1$ must be a double root, which means that $Q'(1) = 0$ as well, in which we have \[ Q'(1) = 2d + \Delta \sum_{j = 1}^{d - 1} 2j - \left( \Delta + \frac{1}{d} \right) \sum_{i = 1}^{d} (2i - 1) = 0 \]Simplifying this gives us $\Delta = 1$, which is what we wanted. Claim 04. The interval $I = \left( 1, 1 + \frac{1}{d} \right)$ works. Proof. Note that for all $x \in \mathbb{R}$, we have \[ P(-x) > x^{2d} + (1 + \varepsilon) \sum_{j = 0}^{d - 1} x^{2j} - \left( 1 + \frac{1}{d} - \varepsilon \right) \sum_{i = 1}^{d} x^{2i - 1} \]for any $\varepsilon > 0$. It suffices to prove that for all $x \in \mathbb{R}$, we have \[ \sum_{j = 0}^{d} x^{2j} \ge \left( 1 + \frac{1}{d} \right) \sum_{i = 1}^{d} x^{2i - 1} \]It suffices to prove the above inequality for $x \in \mathbb{R}^+$. Rearranging, we get \[ \frac{1 + x^2 + x^4 + \dots + x^{2d}}{d + 1} \ge \frac{x + x^3 + \dots + x^{2d - 1}}{d} \]and this is indeed true for any $x \in \mathbb{R}^+$. We will prove this by induction on $d$. Indeed, for $d = 1$, this is just the AM-GM inequality $\frac{x^2 + 1}{2} \ge x$. Now, suppose that it is true for $d = k - 1$, which means that \[ (k - 1)(x^{2k - 2} + \dots + x^2 + 1) \ge k(x + x^3 + \dots + x^{2k - 3}) \]We will now prove this for $d = k$, which is equivalent to proving \[ kx^{2k} + x^{2k - 2} + \dots + x^2 + 1 \ge (k + 1)x^{2k - 1} + x^{2k -3} + \dots + x^3 + x \]However, this is straightforward since for all $i \le k - 1$, we have $x^{2i} (x - 1)(x^{2k - 2i - 1} - 1) \ge 0 \implies x^{2k} + x^{2i} \ge x^{2k - 1} + x^{2i + 1}$, and therefore \[ LHS = \sum_{i = 0}^{k - 1} (x^{2k} + x^{2i}) \ge \sum_{i = 0}^{k - 1} (x^{2k - 1} + x^{2i + 1}) = RHS \]Therefore, the original inequality is true for all $d \in \mathbb{N}$.

The answer is $\left(1 , 1 + \frac{1}{d} \right)$. Write $I = (\alpha,\beta)$ with $\alpha < \beta$. As $P(x)$ has even degree (and leading coefficient positive), so the given assertion is equivalent to \begin{align*} P(x) > 0 ~~ \forall ~ x \in \mathbb R, ~ a_{2d-1},\ldots,a_0 \in (\alpha,\beta) \qquad \qquad (1) \end{align*}Observe $\alpha \ge 0$, otherwise $a_0$ negative, $P(0) < 0$. Let polynomial $$Q(x) = x^{2d} + r \left(x^{2d-2} + \cdots + x^2 + 1 \right) - s \left(x^{2d-1} + \cdots + x^3 + x \right)$$Consider the assertion \begin{align*} Q(x) \ge 0 ~ \forall ~ x \in \mathbb R \qquad \qquad (2) \end{align*} Claim: $(1)$ is equivalent to $(2)$. Proof: Suppose $(2)$ isn't true (i.e. some $Q(\lambda) < 0$). We will show $(1)$ isn't true. In $(1)$, take $$ a_{2d-1} = \cdots = a_3 = a_1 = \beta - \epsilon ~~,~~ a_{2d-2} = \cdots = a_2 = a_0 = \alpha + \epsilon ~~,~~ x = -\lambda$$for a $\epsilon \in (0,s-r)$. Observe as $\epsilon \to 0$, value of $P(-\lambda)$ approaches $Q(\lambda)$. So for a very small $\epsilon$, $P(-\lambda) < 0$, contradicting $(1)$. Now suppose $(2)$ is true. We will show $(1)$ is true. $(1)$ is trivially true for $x \ge 0$. If $x < 0$, then note $P(x) > Q(x) \ge 0$, as desired. $\square$ $x=1$ in $(2)$ gives $$1 + \alpha \cdot d - \beta \cdot d \ge 0 \implies 1 \ge d(\beta - \alpha) \implies \beta - \alpha \le \frac{1}{d} $$So length of $I$ is at most $\frac{1}{d}$. Now we show only such $I$ of length $\frac{1}{d}$ is only $\left( 1 , 1 + \frac{1}{d} \right)$. Put $\beta = \alpha + \frac{1}{d}$. Note we only need to worry about $(2)$ when $x \ge 0$, as it is always true for $x \le 0$. In particular $(2)$ is equivalent to \begin{align*} Q(x) \ge 0 ~~ \forall ~ x >0 \qquad \qquad (3) \end{align*}Observe that \begin{align*} Q(x) &= x^{2d} + \alpha(x^{2d-2} + \cdots + x^2 + 1) + \left( \alpha + \frac{1}{d} \right)( x^{2d-1} + \cdots + x^3 + x) \\ &= -\alpha(x^{2d-1} - x^{2d-2} + \cdots + x^3 - x^2 + x-1) + x^{2d} - \frac{x^{2d-1} + \cdots + x^3 + x }{d} \\ &= -\alpha(x-1)(x^{2d-2} + \cdots + x^2 + 1) + \frac{x-1}{d} \left( \frac{x^{2d} - x^{2d-1} }{x-1} + \cdots + \frac{x^{2d} - x^3}{x-1} + \frac{x^{2d} -x}{x-1} \right) \\ &= (x-1) \left( \alpha(x^{2d-1} + \cdots + x^3 + x) \right) \\ & ~~~ + \frac{x-1}{d} \left( x^{2d-1} + \cdots + (x^{2d-1} + \cdots + x^4 + x^3) + (x^{2d-1} + \cdots + x^2 + x) \right) \\ &= (x-1)(\alpha x^{2d-2} + \cdots + \alpha x^2 + \alpha) \\ &~~~ + \frac{x-1}{d} \left( dx^{2d-1} + (d-1)(x^{2d-2} + x^{2d-3}) + \cdots + 2(x^4 + x^3) + (x^2 + x) \right) \end{align*}Consider the polynomial \begin{align*} g(x) &= -\alpha(x^{2d-2} + \cdots + x^2 + 1) + \frac{1}{d} \left( dx^{2d-1} + (d-1)(x^{2d-2} + x^{2d-3}) + \cdots + 2(x^4 + x^3) + (x^2 + x) \right) \\ &=c_{2d-1}x^{2d-1} + \cdots + c_1x + c_0 \end{align*}where $c_{2d-1},\ldots,c_1 , c_0$ are constants. We have shown that $$ Q(x) = g(x)(x-1)$$Hence $(3)$ is equivalent to \begin{align*} g(x) \ge 0 ~~ \forall ~ x \ge 1 \qquad \text{and} \qquad g(x) \le 0 ~~ \forall ~ 0 < x \le 1 \qquad \qquad (4) \end{align*}Note if $(4)$ is true, then we must have $g(1) = 0$ (this follows by looking at graph of $g$, or by continuity). But, \begin{align*} -g(1) = \alpha \cdot d - \frac{1}{d} \left( d + 2(1 + 2 + \cdots (d-1)) \right) = \alpha \cdot d - \frac{d + d(d-1)}{d} = \alpha \cdot d - d \end{align*}This forces $\alpha = 1$. Now we conversely show that for $\alpha = 1$, $(4)$ is indeed true. Call a polynomial $g(x)$ 1-separated if it satisfies $(4)$. Lemma: If $g(x) = c_nx^n + \cdots + c_1x + c_0$ satisfies the following conditions, then it is 1-seperated: $g(1) = 0$, i.e. $c_n + \cdots + c_1 + c_0 = 0$. $c_n + \cdots + c_k \ge 0 ~~ \forall ~ 0 \le k \le n$. First Proof: We use strong induction on $n$, with base case $n=0$ direct. Now we inductive step. We will write $g$ as a sum of two 1-separated polynomials. Note $c_n \ge 0$. Write $$g(x) = c_n(x^n - x^{n-1}) + x^{n-1}(c_n + c_{n-1}) + c_{n-2}x^{n-2} + \cdots + c_1x + c_0$$The polynomial $c(x^n -x^{n-1})$ is clearly 1-separated. The leftover polynomial is separated because of the induction hypothesis. $\square$ Second Proof: Just write \begin{align*} g(x) &= c_n(x^n - x^{n-1}) + (c_n + c_{n-1})(x^{n-1} - x^{n-2}) + (c_n + c_{n-1} + c_{n-2})(x^{n-2} - x^{n-3}) + \\ &~~~ \cdots + (c_n + \cdots +c_3+c_2)(x^2 - x) + (c_n + \cdots + c_2+c_1)(x- 1) \end{align*}Since each individual polynomial is clearly 1-separated, so $g$ is sum of 1-separated polynomials, as desired. $\square$ Now it suffices to verify that $g$ of the problem satisfies the Lemma conditions. Recall that \begin{align*} &c_{2d-1} = 1, c_{2d-2} = \frac{d-1}{d} - 1, c_{2d-3} = \frac{d-1}{d}, c_{2d-4} = \frac{d-2}{d} - 1, c_{2d-5} = \frac{d-2}{d} , \\& \ldots,c_2 = \frac{1}{d} - 1 , c_1 = \frac{1}{d},c_0 = -1 \end{align*}$g(1) = 0$ is already shown (in fact, $\alpha$ was chosen such that $g(1) = 0$). Now we just have to show \begin{align} S_k := c_{2d-1} + \cdots + c_k \ge 0 ~~~ \forall ~ 0 \le k \le 2d-1 \qquad \qquad (5) \end{align}Basically for any $0 \le t \le d-1$ we have $$c_{2d-1-2t} = \frac{d-t}{d} ~,~ c_{2d-2-2t} = \frac{d-t-1}{d} - 1 ~,~ c_{2d-1-t} +c_t = 0 $$Using $c_0 + \cdots + c_{2d-1} = 0$ we also have $$S_k = -(c_0 + \cdots + c_{k-1}) = c_{2d-1} + \cdots + c_{2d-k} = S_{2d-k} $$So it suffices to show $S_{2d-1},\ldots,S_d \ge 0$. Observe, $$c_{2d-1-2t} + c_{2d-2-2t} = \frac{d-t}{d} + \left( \frac{d-t-1}{d} - 1 \right) = \frac{d-2t-1}{d} \ge 0 ~~ \forall ~ 0 \le t < \frac{d}{2} $$So for any $d \le k \le 2d-1$, we can prove $S_k \ge 0$ by pairing terms (the only term which possible might not get paired will be positive). This completes the proof. $\blacksquare$

My solution Let the open intervals be $(a,b)$ for $a<b$ First note that since $degP=2d$ is even, then $P(x)$ has no real roots iff $P(x)>0$ for all $x \in \mathbb{R}$ (Otherwise there exists $c$ such $P(c)<0$, but then because $\mathop {\lim }\limits_{x \to + \infty } P(x) = + \infty $ and leads to there exists a root in $(c, +\infty)$) Claim 1: $a>0$ Proof Since $P(0)=a_0>0$ then $a>0$ (Otherwise if $a<0$ then we can choose $a_0 \in (a,0)$) Claim 2: $b-a= \frac{1}{d}$ Proof Since \[P( - 1) = 1 + \sum\limits_{2 \mid k} {{a_k}{x^k}} - \sum\limits_{ \text{ k odd} } {{a_k}{x^k}} >0 \] Let $a_{2k-2}=a+ \varepsilon$ and $a_{2k-1}=b- \varepsilon$ for $k=1,2, \ldots, d$, we have $$ P(-1)=1+(a+ \varepsilon ) \cdot d - (b - \varepsilon ) \cdot d =1+d(a-b) +2d \cdot \varepsilon $$ Take $ \varepsilon $ tends to $0$ we have $1+d(a-b) \ge 0 \implies b-a \le \frac{1}{d}$ Now all the open intervals of maximum length will have the form $(a,a+\frac{1}{d})$ Let $a_k=a+f(k)$ for $f(k) \in (0, \frac{1}{d} )$, then $P(x)$ becomes \[\begin{array}{l} P(x) = {x^{2d}} + \sum\limits_{k = 0}^{2d - 1} {{a_k}{x^k}} = {x^{2d}} + \sum\limits_{k = 0}^{2d - 1} {(a + f(k)){x^k}} = {x^{2d}} + a\sum\limits_{k = 0}^{2d - 1} {{x^k}} + \sum\limits_{k = 0}^{2d - 1} {f(k){x^k}} \\ \\ = {x^{2d}} + a\frac{{{x^{2d}} - 1}}{{x - 1}} + \sum\limits_{k = 0}^{2d - 1} {f(k){x^k}} = \frac{{{x^{2d + 1}} + (a - 1){x^{2d}} + (x - 1)\sum\limits_{k = 0}^{2d - 1} {f(k){x^k}} }}{{x - 1}} \end{array}\] Now we are going to prove that $a=1$. Suppose that $a \ne 1$ We have \[\begin{array}{l} \mathop {\lim }\limits_{x \to {1^ - }} ({x^{2d + 1}} + (a - 1){x^{2d}}) = a - 1 + 1 = a \in (0,1) \cup (1,+\infty) \\ \\ \mathop {\lim }\limits_{x \to {1^ - }} (x - 1)\sum\limits_{k = 0}^{2d - 1} {f(k){x^k}} = 0 \end{array}\] Hence there exists $x_0 \in (0,1)$ such that ${{x_0^{2d + 1}} + (a - 1){x_0^{2d}} + (x_0 - 1)\sum\limits_{k = 0}^{2d - 1} {f(k){x_0^k}} } >0$, but since $x_0-1<0$, we have $P(x_0)<0$, which is a contradiction Hence $a=1$. When $a=1$, we have \[P(x) = {x^{2d}} + \sum\limits_{k = 0}^{2d - 1} {{x^k}} + \sum\limits_{k = 0}^{2d - 1} {f(k){x^k}} > 0\forall x > 0\] So we have to prove for $x<0$, $P(x)>0$ We have for $x<0$, \[P(x) > {x^{2d}} + {x^{2d - 2}} + ... + {x^2} + 1 - (1 + \frac{1}{d})({|x|^{2d - 1}} + {|x|^{2d - 3}} + ... + |x|)\] But since ${x^{2d - 2k}} + {x^{2d - 2k - 2}} \ge 2\sqrt {{x^{2d - 2k}}{x^{2d - 2k - 2}}} = 2|x{|^{2d - 2k - 1}}$ by AM-GM, applying here we have \[2({x^{2d}} + {x^{2d - 2}} + ... + {x^2} + 1) \ge {x^{2d}} + 1 + 2\sum\limits_{k = 0}^{d - 1} {|x{|^{2k + 1}}} \]So we have to prove that \[{x^{2d}} + 1 \ge \frac{2}{d}\sum\limits_{k = 0}^{d - 1} {|x{|^{2k + 1}}} \Leftrightarrow d({x^{2d}} + 1) \ge 2\sum\limits_{k = 0}^{d - 1} {|x{|^{2k + 1}}} \]But notice that \[{x^{2d}} + 1 \ge |x{|^{2k + 1}} + |x{|^{2d - (2k + 1)}} \Leftrightarrow (|x{|^{2k + 1}} - 1)(|x{|^{2d - 2k - 1}} - 1) \ge 0 \text{ (Always Right)} \]Which leads to our conclusion by grouping pairs of $|x{|^{2k + 1}}$,$ |x{|^{2d - (2k + 1)}}$