Inequality questions

Question

$$K=\frac{B}{A-B} $$ $$A>B>0$$

Is it possible to prove mathematically that as I increase A and B (by different numbers), while keeping A>B, K will also increase? My thought is as we increase both A and B, A-B will get closer to 0, which will increase K.

Answer 1

$K$ needs not increase because you can take $A = 2B \implies K = \dfrac{1}{2}$ a constant. And to see $K$ could even be decreasing, you can take $A = B^2\implies K = \dfrac{B}{B^2-B} = \dfrac{1}{B-1}, B > 1$ clearly decreases as $B$ increases. And to show the possibility for $K$ to increase, simply take $A = B+1\implies K = B$ will increase with $B$.

Answer 2

let $f(x) = {1 \over x-1}$ and note that for $x>1$ that $f$ is decreasing.

You have $K = f({A \over B})$. Note that I can increase both $A,B$ (while maintaining $A > B$) and have the value of ${A \over B}$ increase or decrease. Hence $K$ can increase or decrease.

If ${A \over B}$ increases then $K$ will decrease.

Answer 3

We can use the total differential to check this :

$$K=\frac B {A-B}$$

$$dK = \frac {\partial K}{\partial A} dA + \frac{\partial K}{\partial B}dB$$

$$dK = \left(\frac {-B}{(A-B)^2}\right){}dA + \left( \frac 1{A-B} + \frac{B}{(A-B)^2}\right)dB$$

$$dK = \left({\frac {-B}{(A-B)^2}}\right){}dA + \left(\frac{A}{(A-B)^2}\right)dB$$

$$dK = \left( \frac 1 {(A-B)^2}\right)\left( -BdA+AdB \right)$$

So we have three possibilities :

• $dK > 0$ This requires $AdB > BdA$

• $dK=0$ This requires $AdB=BdA$

• $dK < 0$ This requires $AdB < BdA$

My thought is as we increase both A and B, A-B will get closer to 0, which will increase K.

So this is not generally true. The effect will depend on the relative values of $A$, $B$ and the changes in those values. The condition $A>B$ does not change this.