# How to prove this equality in proportional fairness analysis?

How to prove: $$\sum_{s=1}^S\left(\frac{y_s-x_s}{x_s}\right)=\bigtriangledown J_\vec x\cdot(\vec y-\vec x)$$ with: $$J_\vec x=\sum_s\ln(x_s)$$

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With $$J_x = \sum_s \ln(x_s)$$ you have \begin{align} \nabla J_x &= (\frac{\partial}{\partial x_1} (\sum_s \ln(x_s)),\frac{\partial}{\partial x_2} (\sum_s \ln(x_s)), ...\frac{\partial}{\partial x_S} (\sum_s \ln(x_s)) \bigr)^T \\&=( \frac{1}{x_1},\frac{1}{x_2},...,\frac{1}{x_S})^T \end{align} furthermore \begin{align} \vec x-\vec y = (x_1-y_1, x_2-y_2,...,x_S-y_S)^T \end{align} If you now take the dot product of the two terms you get \begin{align} \nabla J_x \cdot (\vec x-\vec y) & = ( \frac{1}{x_1},\frac{1}{x_2},...,\frac{1}{x_S})^T \cdot (x_1-y_1, x_2-y_2,...,x_S-y_S)^T \\ & =\frac{x_1-y_1}{x_1} + \frac{x_2-y_2}{x_2}+ ... +\frac{x_S-y_S}{x_S}\end{align} Which is your left hand side.