Northeastern WOOTers Mock AIME I Problems/Problem 5

Problem 5

Let $x$, $y$, and $z$ be real numbers. Given that $x^2+2y^2+3z^2=1$, the maximum value of $\left( 3x+2y+z \right)^2$ can be represented $\frac{m}{n}$, where $m$ and $n$ are positive integers, where $m$ and $n$ are relatively prime. Find $m+n$.



Solution

By the Cauchy-Schwarz Inequality, \[(x^2 + 2y^2 + 3z^2)(9 + 2 + \frac{1}{3}) \geq \left( 3x + 2y + z \right)^2.\] It follows that $\left( 3x + 2y + z \right)^2 \leq \frac{34}{3}$. Since this is attainable when $\frac{x^2}{9} = \frac{2y^2}{2} = \frac{3z^2}{1/3} \Rightarrow x = \frac{9}{\sqrt{102}}, y = \frac{3}{\sqrt{102}}, z = \frac{1}{\sqrt{102}}$, the answer is $34 + 3 = \boxed{037}$.

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