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Let $x_1,\ldots x_n,y_1,\ldots y_n$ be reals and $\bar{x},\bar{y}$ the aritmetic mean of numbers $x_1,\ldots x_n$ and $y_1,\ldots y_n$ respectively. How can I show that $$-1\leq \dfrac{\sum_{i=1}^n (x_i-\bar{x})(y_1-\bar{y})}{\sqrt{\sum_{i=1}^n(x_i-\bar{x})^2\sum_{i=1}^n(y_i-\bar{y})^2}}$$ Looks a bit like Cauchy (it gives me that the expression is at most one) but I was unable to find the proof.

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You're right, it is a variant of the Cauchy inequality. – Raskolnikov Oct 9 '12 at 22:05
up vote 1 down vote accepted

It's just the Cauchy--Schwarz inequality applied to the two vectors $(x_1-\bar x,\ldots,x_n - \bar x)$ and $(y_1-\bar y,\ldots,y_n - \bar y)$.

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Is just the relationship between a scalar product (numerator), the norm of the vectors (denominator) and the cosinus of the angle.

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