Denote $P(x) = 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x^1 + 2$. Since $c_i \in (0,2)$ for all $1\leq i \leq n-1$, it follows $P(x) \geq 2 > 0$ for all $x\leq 0$. It remains to prove $P(x) > 0$ also for all $x> 0$. Let $c_i = 1+\epsilon_i$, so $\displaystyle \sum_{i=1}^{n-1} \left\lvert \epsilon_i \right\rvert < 1$. Then for $x> 0$ we have \[P(x) = (x^n - x^{n-1} + x^{n-2} - \cdots - x + 1) + x^{n} + 1 + \sum_{i=1}^{n-1}(-1)^i \epsilon_ix^i.\] We have $x^n - x^{n-1} + x^{n-2} - \cdots - x + 1 = \dfrac {x^{n+1} + 1}{x+1} > \dfrac {1} {2} > 0$ since $2x^{n+1} - x + 1 > 0$ (for $x\geq 1$ we have $x^{n+1} - x \geq 0$ while for $x\leq 1$ we have $1 - x \geq 0$). For $0 < x \leq 1$ we have $\displaystyle \left | \sum_{i=1}^{n-1}(-1)^i \epsilon_ix^i \right | \leq \sum_{i=1}^{n-1} \left |(-1)^i \epsilon_ix^i\right | \leq \sum_{i=1}^{n-1} \left |\epsilon_i\right | < 1$, and so $\displaystyle x^{n} + 1 + \sum_{i=1}^{n-1}(-1)^i \epsilon_ix^i \geq x^{n} + 1 - \left | \sum_{i=1}^{n-1}(-1)^i \epsilon_ix^i \right | > x^n > 0$. For $1 \leq x$ we have $\displaystyle \left | \sum_{i=1}^{n-1}(-1)^i \epsilon_i(1/x)^{n-i} \right | \leq $ $\sum_{i=1}^{n-1} \left |(-1)^i \epsilon_i(1/x)^{n-i}\right | \leq$ $ \sum_{i=1}^{n-1} \left |\epsilon_i\right | < 1$, and so $\displaystyle x^{n} + 1 + \sum_{i=1}^{n-1}(-1)^i \epsilon_ix^i =$ $ x^{n}\left (1 + (1/x)^n + \sum_{i=1}^{n-1}(-1)^i \epsilon_i(1/x)^{n-i}\right ) \geq$ $ x^n\left (1 + (1/x)^n - \left | \sum_{i=1}^{n-1}(-1)^i \epsilon_i(1/x)^{n-i} \right |\right ) > 1 > 0$. Thus $P(x) > \dfrac {1} {2} > 0$, with room to spare, so we may either relax the condition $\displaystyle \sum_{i=1}^{n-1} \left\lvert c_i-1 \right\rvert < 1$ or lower the free term of $P(x)$ to $\dfrac {3}{2}$.

Thank you, @mavropnevma, very beautiful proof! can we use since: \[ P(x+1)(x+1)=2x^n+(2-c_{n-1})x^n-(c_{n-1}-c_{n-2})x^{n-1}+(c_{n-2}-c_{n-3})x^{n-2}+...+(c_2-c_1)x^2+(2-c_1)x+2=0 ? \]

Let $ a_i = (-1)^i(1 - c_i) $ for all integers $ 1 \le i \le n - 1 $. Thus the condition becomes $ \sum_{i = 1}^{n - 1}|a_i| < 1 $ and our polynomial can be written as $ P(x) = x^n + 1 + \frac{x^{n + 1} + 1}{x + 1} - \sum_{i = 1}^{n - 1}a_ix^i $ (when $ x \ne -1 $). Now we proceed with some casework: Case 1: $ -1 < x \le 1 $ Here we have $ \sum_{i = 1}^{n - 1}a_ix^i \le \sum_{i = 1}^{n - 1}|a_i| < 1 < x^n + 1 + \frac{x^{n + 1} + 1}{x + 1} $ as desired. Case 2: $ x > 1 $ or $ x < -1 $ Here we have $ \sum_{i = 1}^{n - 1}a_ix^i < |x|^{n - 1}\sum_{i = 1}^{n - 1}|a_i| < |x|^{n - 1} < 1 + x^n + \frac{x^{n + 1} + 1}{x + 1} $ as desired. Case 3: $ x = -1 $ Here we have that $ P(-1) = n + 3 - \sum_{i = 1}^{n - 1}a_i(-1)^i > n + 3 - \sum_{i = 1}^{n - 1}|a_i| > n + 2 > 0 $ as desired. Therefore $ P(x) > 0 $ for all $ x \in \mathbb{R} $ which implies the desired result.

Note that $c_i>0$ for all $i$. Thus if $x<0$, clearly the polynomial (call it $P(x)$) is positive. Also, note that we can switch $x$ and $\frac{1}{x}$ without switching the problem statement. So WLOG assume that $x\ge 1$. Let $d_i=c_i-1$, so $|d_1|+|d_2|+\cdots+|d_{n-1}|<1$. Then \begin{align*} |P(x)|&=|2x^n-c_{n-1}x^{n-1}+c_{n-2}x^{n-2}-\cdots-c_1x^1+2|\\ &=|2x^n-x^{n-1}+x^{n-2}-\cdots-x^1+2-d_{n-1}x^{n-1}+d_{n-2}x^{n-2}+\cdots-d_1x^1|\\ &\ge |2x^n-x^{n-1}+x^{n-2}-\cdots-x^1+2|-|d_{n-1}x^{n-1}+d_{n-2}x^{n-2}+\cdots-d_1x^1|\\ &\ge |2x^n-x^{n-1}+x^{n-2}-\cdots-x^1+2|-(|d_{n-1}x^{n-1}|+|d_{n-2}x^{n-2}|+\cdots+|d_1x^1|)\\ &\ge |2x^n-x^{n-1}+x^{n-2}-\cdots-x^1+2|-(|d_{n-1}|+|d_{n-2}|+\cdots+|d_1|)x^{n-1}\\ &> |2x^n-x^{n-1}+x^{n-2}-\cdots-x^1+2|-x^{n-1}\\ &\ge 2x^n-2x^{n-1}+x^{n-2}-\cdots -x^1+2\\ &= (2x^n-2x^{n-1})+(x^{n-2}-x^{n-3})+\cdots+(x^2-x)+2\\ &\ge 2>0 \end{align*}

We will prove the polynomial is positive for all $x \in {\mathbb R}$. As $c_i > 0$, the result is vacuous for $x \le 0$, so we restrict attention to $x > 0$. Then letting $c_i = 1 - d_i$ for each $i$, the inequality we want to prove becomes \[ x^n + 1 + \frac{x^{n+1}+1}{x+1} > \sum_1^{n-1} d_i x^i \qquad\text{given } \sum |d_i| < 1. \]But obviously $x^n+1 > x^i$ for any $1 \le i \le n-1$ and $x > 0$. So in fact $x^n+1 > \sum_1^{n-1} |d_i| x^i$ holds, as needed.

Slightly different from above, though similar in spirit. Let $f(x)$ be the polynomial in question, and let \[ g(x) := 2x^n - x^{n-1} + x^{n-2} - \cdots - x + 2 = x^n + 1 + \frac{x^{n+1} + 1}{x+1}. \]Suppose $x$ satisfies $f(x) = 0$, and observe that $x > 0$ since the coefficients of $f$ alternate in sign. Remark that \begin{align*} |g(x)| &= |g(x) - f(x)| = |(c_{n-1} - 1)x^{n-1} - (c_{n-2} - 1)x^{n-2} + \cdots + (c_1 - 1)x|\\ &\leq |c_{n-1} - 1|x^{n-1} + |c_{n-2} - 1|x^{n-2} + \cdots + |c_1 - 1|x \leq \max(x,x^{n-1}), \end{align*}where in the last step we use the assumption $\textstyle\sum_{i=1}^{n-1}|c_i - 1| < 1$. That is, any root of $f$ must also satisfy \[ x^n + 1 < g(x) < \max(x,x^{n-1}). \]But this clearly cannot hold: if $x \leq 1$ then the right hand side is at most $1$, while if $x > 1$ then $x^{n-1} < x^n$. So no real root can exist. $\blacksquare$ Remark. Intuitively, this result seems plausible because the coefficients of $f$ are "close" to those of $g$, which has no real roots. This means we're "perturbing" the polynomial $g$ by only a small amount; since the roots of polynomials should vary continuously as the coefficients vary continuously, we expect the roots of $f$ to be "close" to the roots of $g$. In fact, this result is true in general. It is known that, for any polynomial $f$ and for any $\varepsilon > 0$ there exists a $\delta > 0$ such that, if the coefficients of $g$ differ from the coefficients of $f$ by at most $\delta$, the roots of $g$ differ from the roots of $f$ by at most $\varepsilon$. In the case where the roots are distinct, this follows as an application of Implicit Function Theorem, since $f'(t) \neq 0$ whenever $t$ is a simple root of $f$. But more care is needed in the general case.

Let \[P(x) = 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x^1 + 2.\] Lemma: $c_i>0$ for all $i$. Proof: If $c_k \le 0$ for some $k$, then $|c_k-1| > 1,$ contradicton. $\square$ Thus it follows that for $x\le 0$ then $P(x)>0$. Let $c_i = 1- z_i.$ Note that $\sum|z_i| < 1.$ It suffices to show $P(x) > x$ for all $x \in \mathbb{R}.$ Claim: $P(x) > 0$ for all $x \in \mathbb{R}.$ Proof: We have already shown that nonpositive $x$ give the polynomial a positive value. Also, $P(0) =2.$ Using the value of $c_i=1-z_i,$ we can write $P(x)$ as \begin{align*} P(x) &= 2x^n - (1-z_1)x^{n-1} + (1-z_2)x^{n-2}+\cdots+2\\ &= x^n+(x^n-x^{n-1}+x^{n-2}+\cdots-x + 1) + 1 -\sum_{k=1}^{n-1}z_kx^k.\\ &= x^n + 1 + \frac{x^{n+1}+1}{x+1}-\sum_{k=1}^{n-1}z_kx^k. \end{align*}We want to show \[x^n+1+\frac{x^{n+1}+1}{x+1} > \sum_{k=1}^{n-1}z_kx^k.\]We know that $x^n + 1 > x^t$ for $1\le t \in \mathbb{Z}^{+}$ and $x$ positive. Thus, we can see $x^n+1 > \sum_{i=1}^{n-1} |z_i| x^i.$ $\blacksquare$

Does this work? It seems different from other solutions on the thread. Let $P(x)$ be the polynomial. It suffices to show that $P(x) > 0$ for all real $x$. Clearly all the $c_i$ are positive, so $P(x) > 0$ for $x\le 0$. We proceed with $2$ cases for positive $x$: If $x <1$, then $2x^n + \sum_{i=1}^{\frac{n}{2}-1}\left(c_{2i}x^{2i} - c_{2i+1}x^{2i+1}\right) + 2-c_1x$ $\ge 2x^n + \sum_{i=1}^{\frac{n}{2}-1}(c_{2i} - c_{2i+1})x^{2i} + 2 - c_1$ $\ge 2x^n - \sum_{i=1}^{\frac{n}{2}-1}|c_{2i} - c_{2i+1}|x^{2i} + (1-c_1) + 1$ $\ge 2x^n - \sum_{i=1}^{\frac{n}{2}-1}|c_{2i} - c_{2i+1}| + (1-c_1) + 1$ $\ge 2x^n - \sum_{i=1}^n |c_i-1| + 1$ $ > 0$. If $x\ge 1$, then $2 + \sum_{i=1}^{\frac{n}{2}-1}\left(c_{2i}x^{2i} - c_{2i-1}x^{2i-1}\right) + 2x^n - c_{n-1}x^{n-1}$ $\ge 2 + \sum_{i=1}^{\frac{n}{2}-1}(c_{2i} - c_{2i-1})x^{2i} + 2x^n - c_{n-1}x^n$ $\ge 2 - \sum_{i=1}^{\frac{n}{2}-1}|c_{2i} - c_{2i-1}|x^{2i} + (1-c_1)x^n + x^n$ $\ge 2 - \sum_{i=1}^{\frac{n}{2}-1}|c_{2i} - c_{2i-1}|x^n + (1-c_1)x^n + x^n$ $\ge 2 + (1 - \sum_{i=1}^n |c_i-1|)x^n$ $ > 0$ and we are done.

MOHS rating ladder 5M Note that $0<c_i<2$ for any $1\le i\le n-1$. Let\[f(x)=2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x^1 + 2\]We will prove that for any real $x$, $f(x)$ is positive. Case 1: $x\le 0$. If $x=0$, then $f(x)=2$. If $x<0$, then $x^{n-odd}<0$ and $x^{n-even}>0$. The result follows since $c_i>0$ for any $i$ (because we are adding positive times positive and subtracting negative times positive). $\blacksquare$ Case 2: $x>0$. Clearly changing any $c_i$ to $2-c_i$ doesn't change the summation condition. To minimize $f(x)$, we set all $c_i\ge 1$ for $i$ odd and $c_i\le 1$ for $i$ even. Let $c_i=1+a_i$ for odd $i$ and $1-a_i$ for even $i$. Note that $a_i\ge 0$ for all $i$. We have\[f(x)=2x^n-x^{n-1}+x^{n-2}-\cdots+x^2-x+2-\sum_{i=1}^{n-1}a_ix^i\]and\[\sum_{i=1}^{n-1}a_i<1.\]It suffices to show that\[2x^n-x^{n-1}+x^{n-2}-\cdots+x^2-x+2> \sum_{i=1}^{n-1}a_i x^i\]Note that $x^n-x^{n-1}+\cdots-x+1=\frac{x^{n+1}+1}{x+1}$. So we want to show\[x^n+1+\frac{x^{n+1}+1}{x+1}> \sum_{i=1}^{n-1}a_i x^i\]However, we have $x^n+1\ge x^i$ for any $1\le i\le n-1$, so since $\sum a_i<1$, we have\[x^n+1>\sum_{i=1}^{n-1}a_i x^i,\]which is sufficient because $\frac{x^{n+1}+1}{x+1}>0$ $\blacksquare$

5 minute solve. Denote $P(x) = 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x^1 + 2$. It is enough to show that $P(x) > 0$ for all $x$. Writing $c_i=1+\epsilon_i$, our condition becomes $\sum_{i=1}^{n-1} \left\lvert \epsilon_i \right\rvert < 1$. We have $P(x) = (x^n-x^{n-1}+x^{n-2}-...-x+1)+(x^n-\epsilon_{n-1}x^{n-1}+\epsilon_{n-2}x^{n-2}-...-\epsilon_1x+1) = \frac{x^{n+1}+1}{x+1} + G(x)$. Since $\frac{x^{n+1}+1}{x+1} > 0$, it is suffices to show $G(x)>0$. $G(x) = (x^n+1)-(\epsilon_{n-1}x^{n-1}-\epsilon_{n-2}x^{n-2}+...\epsilon_{1}x) = (x^n+1)-R(x)$. It is easy to see that when $x \le 0$, $G(x)>0$ due to $n$ being even, so all that is left to consider is $x>0$. We now take three cases. $0<x<1$ : We have $x^n+1>1$, and observe that $\max(R(x))$ is obtained when $\epsilon_1=1$. We now have $G(x) = 1-x+x^n$, and since $x<1$, $G(x)>0$. $x=1$ : $G(1) = 2-(\epsilon_{n-1}-\epsilon_{n-2}+...+\epsilon_{1})$. By our condition, $G(1) \ge 2 - \sum_{i=1}^{n-1} \left\lvert \epsilon_i \right\rvert$, so $G(x)>0$. $x>1$ : Observe that $\max(R(x))$ is obtained when $\epsilon_{n-1}=1$, so we have $G(x) = x^n-x^{n-1}+1$. Since $x^n>x^{n-1}$, we have $G(x)>0$. Since we have shown that $G(x)>0$ for all $x$, we are done. $\blacksquare$

Here is my solution: https://calimath.org/pdf/USATST2014-4.pdf And I uploaded the solution with motivation to: https://youtu.be/CXHGAfc-3u4

oops took 40 min to solve this in school Let $d_i=c_i-1$. Clearly, there are no roots when $x$ is negative since all of the terms would be positive. Assume $x$ nonnegative for the rest of the solution. We will claim that $$2x^n+2> c_{n-1}x^{n-1}-c_{n-2}x^{n-2}\cdots+ c_1x.$$This can be rearranged as $$2x^n+2> d_{n-1}x^{n-1}-d_{n-2}x^{n-2}\cdots+ d_1x + (x^{n-1}-x^{n-2}\cdots +x)$$$$2x^n+2> d_{n-1}x^{n-1}-d_{n-2}x^{n-2}\cdots+ d_1x + \frac{x^n+x}{x+1}.$$ We will then take cases. Note the constraint $$|d_1|+|d_2|\cdots +|d_{n-1}|<1$$ Case 1: $x=0$. Then we have $2>0$ so this works. Case 2: $0<x<1.$ Then, in order to maximize $$d_{n-1}x^{n-1}-d_{n-2}x^{n-2}\cdots+ d_1x + \frac{x^n+x}{x+1},$$we would put all of the weight on $x$ since it is the largest out of the powers of $x$ (in other words, $d_1=1$ and all other $d_i$ are 0 (we let the condition on $d$ be nonstrict and show a more general result)). This means that it suffices to show $$2x^n+2>x+\frac{x^n+x}{x+1}.$$This is true since we have $$2x^n>0,1>x,$$$$1=\frac{x+1}{x+1}>\frac{x^n+x}{x+1}.$$ Case 3: $x=1$. Then, we need to show $$4>d_{n-1}-d_{n-2}\cdots +d_1 +1,$$which is clearly true due to the constraint $|d_1|+|d_2|\cdots +|d_{n-1}|<1$. Case 4: $x>1.$ Then, in order to maximize $$d_{n-1}x^{n-1}-d_{n-2}x^{n-2}\cdots+ d_1x + \frac{x^n+x}{x+1},$$we want to instead put all of the weight on $x^{n-1}$ since it is the largest. Therefore, it suffices to show that $$2x^n+2>x^{n-1}+\frac{x^n+x}{x+1}.$$This follows from the inequalities $$x^n>x^{n-1},$$$$x^n>\frac{x^n+x}{2}>\frac{x^n+x}{x+1},$$$$2>0.$$ Since we have covered every case, we are done. edit: this is apparently 0 MOHS??? what. mind = blown

huh why is my solution so simple compared to everyone else's? did i do something wrong Suppose that there is a real root, say $x$. We will casework on $x$. Case 1. $x\le 0$. If $x<0$ then the expression is positive, if $x=0$ then the expression is $2$(which is not $0$). Case 2. $0<x<1$. Dividing the polynomial by $x^n$ gives a polynomial in $\frac1x$ which will work in the next case. Case 3. $1\le x$. Group the terms of the polynomial as follows: \begin{align*} &x^n+2+(1\cdot x-c_{n-1})x^{n-1}+(c_{n-2}\cdot x-c_{n-3})x^{n-3}+\dots+(c_2\cdot x-c_1)x \\ \ge &x^n+2+(1-c_{n-1})x^{n-1}+(c_{n-2}-c_{n-3})x^{n-3}+\dots+(c_2-c_1)x. \end{align*}I claim that the second expression is always positive. Therefore, we will try to make everything as small as possible, so we want each of the expressions in parentheses to be nonpositive(we can always do this since it's better for the economy if $c_{n-1}$ is $1$ rather than less than $1$, and everything else you can just swap the two numbers). Therefore, \begin{align*} &x^n+2+(1-c_{n-1})x^{n-1}+(c_{n-2}-c_{n-3})x^{n-3}+\dots+(c_2-c_1)x \\ \ge &x^n+2+(1-c_{n-1}+c_{n-2}-c_{n-3}+\dots+c_2-c_1)x^n \\ \ge &x^n+2+(-1)x^n \\ = &2\;\blacksquare \end{align*}

sorry this is 0M? this took me longer than 2023 IMO 5 by at least 1.5 hours i guess this is a testament to how bad i am at alg As an observation, since $c_i > 0$, the coefficients are alternating in sign so nonpositive $x$ will never work (also note $x=0$ provides $2>0$). Henceforth assume $x > 0$. Now let $a_i = c_i - 1$, and split the polynomial into two polynomials: \[ 2x^n - c_{n-1}x^{n-1} + \cdots + 2 = (x^n - x^{n-1} + \cdots + 1) + x^n - a_{n-1}x^{n-1} + \cdots + 1.\] The first polynomial is always positive since \[ \frac{x^{n+1} - 1}{x - 1} \]is always positive. $x=1$ can be tested separately. Now the condition rewrites $\sum |a_i| < 1$. Hence the negative coefficients in absolute value of the second polynomial sum to less than 1. If $x > 1$, then change every term with a negative coefficient to be degree $n$. This decreases the value, but since the negative coefficients in absolute value sum to less than $1$, we have an fail since $x^n$ exists. If $0 < x < 1$, then change every term with a negative coefficient to be degree $0$. This again decreases the value and by basically the same exact argument we have another fail since $1$ exists.