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Find the smallest possible value of $\omega$ such that the system of inequalities \begin{align*} (1009\cdot1010-2018)x+2018\leq2018\omega\\ (1008\cdot{1009}-2018)x+2018\leq2016\omega \\ (1007\cdot{1008}-2018)x+2018\leq2014\omega \\ .................................... \\ (1\cdot2-2018)x+2018\leq2\omega \\ \end{align*}

has a real solution in $x$.

Subtracting the first inequality from the second inequality, we get that $x=\frac{1009}{2017}$. This is also correct when we subtract the second inequality from the third one. When I plugged in $x$, I got an unruly fractional value for $\omega$. Can someone guide me through the solution? Help is much appreciated.

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You have $$(n(n+1)-2018)x+2018\le 2nw$$ for $n=1,2,\dots,1009$.

The solution to one of these is $$x\begin{cases}\le\frac{2nw-2018}{n(n+1)-2018}&n\ge45\\\ge\frac{2nw-2018}{2018-n(n+1)}&n\le44\end{cases}$$ Where we get that $44$ because the positive root of the quadratic on the denominator is $\alpha=\frac{-1+3\sqrt{897}}{2}\approx 44.4$

You want there to exist an $x$ for which all of these inequalities hold. To do this, let's work out for which $n$ the first fraction is smallest, and for which $n$ the second fraction is largest.

It turns out that the fractions have larger magnitude for larger $n$. If $w > \alpha/2$, then the maximum of the second fraction occurs at $n=44$. If $w<\alpha/2$ then the fraction is always negative on the range $n\le44$, and the maximum occurs at $n=1$. Similarly for the other fraction, if $w>\alpha/2$, the minimum occurs at $n=1009$, and if $w>\alpha/2$ then the minimum occurs at $n=45$. (I worked this all out by looking at the graphs of the fraction, with $n$ as the variable, and by varying $w$.)

Let's first take the case $w<\alpha/2$. Then we need $$\frac{2w-2018}{2016}\le x\le\frac{90w-2018}{52}$$As before, we equate these two sides to give $52(2w-2018)=2016(90w-2018)$. This give $181336 w=3963352\implies \bf{w=21.8564}\cdots$, which satisfies $w<\alpha/2$, so is a valid solution. This gives the real solution of $x=-0.9793091\cdots$

The other case cannot produce a smaller solution, so this is the smallest possible solution.

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