Solving the inequality between a and b [closed]

I run into this inequality

$$(a + b)^{1 - \epsilon} \;a < b$$

where $$a \in \mathbb{Z}^+$$ and $$\epsilon \in (0, 1)$$. What value (w.r.t $$a$$ and $$\epsilon$$) should I set $$b$$ equal to such that this inequality holds for all $$a \in \mathbb{Z}^+$$? (if possible).

What I have gotten so far:

$$\frac{a + b}{(a + b)^{\epsilon}} \; a < b$$ $$\Downarrow$$ $$a^2 + ab < (a + b)^{\epsilon} \; b \leq (a^{\epsilon} + b^{\epsilon}) b$$ $$\Downarrow$$ $$a^2 + ab < a^{\epsilon}b + b^{1 + \epsilon}$$

I am struggling to proceed from here.

• are you hoping that the inequality will hold for all $a, b \in \mathbb{Z}^+$? take $\epsilon=1/2$, $a=2$, $b=1$ for a counterexample. – Carlo Beenakker Apr 21 at 14:53
• I guess my question is, suppose this inequality holds, what is the relation between a and b. – Null_Space Apr 21 at 14:55
• The inequality holds for $\epsilon>1-\ln(b/a)/\ln(a+b)$. In particular, it will certainly fail when $\epsilon$ is very small. – Neil Strickland Apr 21 at 14:55
• I have restated the question. Thanks! – Null_Space Apr 21 at 14:59
• Why is this question labeled "off-topic"? – Null_Space Apr 22 at 0:34

Rewrite the inequality as $$f_{p,a}(b):=\frac b{(a+b)^p}>a,\tag{1}$$ where $$p:=1-\epsilon\in(0,1)$$. We have $$f'_{p,a}(b)=\frac{a+(1-p)b}{(a+b)^{p+1}}>0$$ for $$a,b>0$$, so that the function $$f_{p,a}$$ is continuous and strictly increasing from $$0$$ to $$\infty$$ on $$(0,\infty)$$. So, (1) can be rewritten as $$b>f_{p,a}^{-1}(a),\tag{1}$$ where $$f_{p,a}^{-1}$$ is the function inverse to $$f_{p,a}^{-1}$$.
If $$p$$ is rational, then the function $$f_{p,a}^{-1}$$ is algebraic. Otherwise, it apparently cannot be expressed in closed form; at least, Mathematica cannot do that:
• Thanks a lot! Suppose $p$ is irrational, say $1 / \sqrt{2}$. Let $p' > p$ be a rational number within $(0, 1)$, can we set $b$ to satisfies: $b > f_{p', a}^{-1}(a)$ since it follows that $b > f_{p', a}^{-1}(a) > f_{p, a}^{-1}(a)$? – Null_Space Apr 21 at 15:34
• @Null_Space : No, we do not have the monotonicity in $p$. – Iosif Pinelis Apr 21 at 16:02
• @losif I see. So given an irrational $p$ and some positive integer $a$, we cannot compute the value of $b$ to satisfies the inequality right? – Null_Space Apr 21 at 16:08
• @Null_Space : Of course, since the function $f_{p,a}$ is continuous and strictly increasing from $0$ to $\infty$ on $(0,\infty)$, you can compute the values of $f^{-1}_{p,a}$ numerically with any degree of accuracy, for any given $p$ and $a$. Since $a$ is an integer, the computation problem is even easier than that. The only thing is that we cannot express $f^{-1}_{p,a}$ in closed form. – Iosif Pinelis Apr 21 at 16:13