PolynomialAndEquation Inequality Intermediate
2003


Problem - 3125
Let $x$ be a negative real number. Find the maximum value of $y=x+\frac{4}{x} +2007$.

Show solution
$y=-(-x-\frac{4}{x})+2007 = -(\sqrt{-x}-\frac{2}{\sqrt{-x}})^2 + 2003 \le \boxed{2003}$

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