group subtraction inequality

Question

suppose I have four positive real numbers $a$, $b$, $c$, $d$ and a positive scalar $\alpha$ such that these two equations hold: $$a-\alpha ~b \leq c-\alpha ~d $$ $$a^2-\alpha ~b^2 \geq c^2-\alpha ~d^2 $$ We can find these four positive numbers and a positive scalar. Consider $a=11,b=10, \alpha=1, c=5, d=3$.

Now the question is can we find a positive scalar $\rho$ dependent on $\alpha$ value such that we have: $$a-\rho ~b \geq c-\rho ~d $$

For example, for the previous set of numbers $a=11,b=10, \alpha=1, c=5, d=3$, we can set $\rho=\frac{\alpha}{2}=\frac{1}{2}$.

This question is very important for me to answer. I would appreciate if anybody can help to uncover it.

Answer 1

If you let $\rho$ depend on $a, b, c, d,$ and $\alpha$ then you can get this to work out.
From your second inequality $a^2 - c^2 \geq \alpha (b^2 - d^2)$ and thus $(a-c)(a+c) \geq \alpha (b-d)(b+d)$.
So we get: $a-c \geq \alpha (\frac{b+d}{a+c}) (b-d)$.
From the inequality you are looking for: $a - \rho b \geq c - \rho d$ we get $a-c \geq \rho (b-d)$.
Thus, let $\rho = \alpha (\frac{b+d}{a+c})$

Answer 2

I shall prove that a constant $\rho>0$ that depends only on $\alpha$ does not exist. Suppose on the contrary that $\rho$ exists. Hence, for any $a,b,c,d>0$ such that $a-\alpha b\leq c-\alpha d$ and $a^2-\alpha b^2\geq c^2-\alpha d^2$, it follows that $$\alpha-\rho b \geq c-\rho d\,.$$

First, take $a=2+\frac{1}{\alpha}$, $b=1+\frac{2}{\alpha}$, $c=\frac{1}{\alpha}$, and $d=1$. We see that $$a-\alpha b=\frac{1}{\alpha}-\alpha=c-\alpha d\text{ and }a^2-\alpha b^2 =\frac{1}{\alpha^2}-\alpha=c^2-\alpha d^2\,.$$ Since $\rho(b-d)\leq a-c \leq \alpha (b-d)$ and $b-d>0$, we conclude that $\rho \leq \alpha$.

Since $a:=\frac{1}{\alpha}$, $b:=1$, $c:=2+\frac{1}{\alpha}$, and $d:=1+\frac{2}{\alpha}$ also lead to a solution, we have $b-d<0$ and $\rho(b-d)\leq a-c \leq \alpha(b-d)$. That is, $\rho\geq \alpha$. That means $\rho=\alpha$ is the only possibility.

Hence, if $\rho$ exists, then $a-\alpha b=c-\alpha d$ must follow. However, there are other solutions. As an example, you can take $(a,b,c,d):=\left(\frac{5}{2}+\frac{2}{\alpha},1+\frac{2}{\alpha},\frac{3}{2}+\frac{2}{\alpha},1\right)$, where we have $$a-\alpha b=\frac{1}{2}+\frac{2}{\alpha}-\alpha <\frac{3}{2}+\frac{2}{\alpha}-\alpha =c-\alpha d$$ and $$a^2-\alpha b^2 =\frac{9}{4}+\frac{6}{\alpha}+\frac{4}{\alpha^2}-\alpha=c^2-\alpha d^2\,.$$ This is a contradiction.