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I'm trying to check whether this inequality holds for large values of $B$

$$\frac{-AB^2}{B-x_2}+\frac{2y_2}{B}<\frac{-AB^2}{B-x_1}+\frac{2y_1}{B}$$

where $x_2>x_1$ $B>x_2$, $y_2>y_1$ and all other variables are positive. Does this inequality hold for large values of $B?$


Since the $\frac{-AB^2}{B-x_2}$ denominator is smaller than $\frac{-AB^2}{B-x_1}$ it seems intuitively to hold. However, in the limit (where $B \rightarrow \infty$, the two sides should be equal?). How can you check rigorously that this is satisfied?

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Note that

$$\frac{-AB^2}{B-x_2}+\frac{2y_2}{B}<\frac{-AB^2}{B-x_1}+\frac{2y_1}{B} \iff \frac{-AB}{1-\frac{x_2}B}+\frac{2y_2}{B}<\frac{-AB}{1-\frac{x_1}B}+\frac{2y_1}{B}$$

$$ (-AB)\left(1+\frac{x_2}B\right)+\frac{2y_2}{B}<(-AB)\left(1+\frac{x_1}B\right)+\frac{2y_1}{B}$$

$$ -AB-Ax_2+\frac{2y_2}{B}<-AB-Ax_1+\frac{2y_1}{B}$$

$$\frac1B(2y_2-2y_1)< A(x_2-x_1)\iff B>\frac{2y_2-2y_1}{A(x_2-x_1)}$$

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  • $\begingroup$ I think you have a mistake with the minus sign from $1-x_2/B$ to $1+x_2/B$ (from first to second line) $\endgroup$ – pafnuti Feb 8 '18 at 21:09
  • $\begingroup$ @pafnuti I've applied the binomial approximation for B large, indeed $(1+a)^{-1}\approx 1-a$ for $a\to0$ $\endgroup$ – gimusi Feb 8 '18 at 21:11

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