Given $a,b\in\{1,2,3,4,5,6,7,8,9,10\}$ and $b>\frac{a^4}{a^2+1}$, prove $b\geq a^2$

Given that $$a,b \in \{ 1,2,3,4,5,6,7,8,9,10 \}$$ and $$b> \frac{a^4}{a^2+1}$$ How can I prove $$b\geq a^2$$ since I'm looking for all possible values of $$(a,b)$$ (and I actually know all though by some brute force) ?

So far I can go to the canonical form of the original inequality is this

$$b> a^2-1+ \frac{1}{a^2+1}.$$

Any help will be much appreciated :)

PS: I already solve this the way I wanted and I've seen my mistakes as well. Thanks to all who helped me and who edited my problem especially to @quasi. I'm SO satisfied rn since it's actually part of a more intricate probabilistic problem. I know it's kind of unfair but I'm more comfortable of my own solution and I put it below...

• Thank you @MPW for editing. Aug 21, 2020 at 3:59
• How do you get $b>a^2?$ For example, $a=b=1$ satisfies the first inequality. Aug 21, 2020 at 3:59
• if you take a=b=1 don't you have b=a^2 and your inequality is satisfied? Aug 21, 2020 at 4:00
• $-1+\frac {1} {a^2 +1}$ is a small difference like it won't affect after all (by some brute force). Aug 21, 2020 at 4:03
• Since I want an elegant solution. I want to do it in casework, but how?? Aug 21, 2020 at 4:05

As noted in the comments, you can't prove $$b > a^2$$ since for the case $$a=b=1$$, the inequality $$b > \frac{a^4}{a^2+1}$$ holds but the inequality $$b > a^2$$ fails.

But to prove $$b\ge a^2$$ holds for all the cases, we can argue as follows . . . \begin{align*} &b > \frac{a^4}{a^2+1} \\[4pt] \implies\;& b > \frac{a^4-1}{a^2+1} \\[4pt] \implies\;& b > \frac{(a^2+1)(a^2-1)}{a^2+1} \\[4pt] \implies\;& b > a^2-1 \\[4pt] \implies\;& b \ge (a^2-1)+1\;\;\;\;\text{[since b and a^2-1 are both integers]} \\[4pt] \implies\;& b \ge a^2 \\[4pt] \end{align*}

• Thank you SO SOmuch @quasi your previous comment thought me everything I needed. Though it took me a while to comprehend it. I now have my own solution here please check it. Thanks again. Aug 21, 2020 at 12:13
• @hansduran0123: Yes, your answer is fine. Aug 21, 2020 at 21:46
• Yes. Thanks to you. Aug 22, 2020 at 10:54

Let’s use contradiction. Suppose $$b\leq a^{2}-1$$.

$$b\leq a^{2}-1$$

$$b(a^{2}+1)\leq (a^{2}-1)(a^{2}+1)$$

$$b\leq \frac{a^{4}-1}{a^{2}+1}$$

Contradicting $$b>\frac{a^{4}}{a^{2}+1}$$

• The negation of $b>a^2$ is $b\le a^2$ Aug 21, 2020 at 12:14
• Honestly, I don't know how to use contradiction in this kind of case yet, and never have I thought it can be used here. Thank you for showing me another way of solving it. Probably next time I'm going to learn this. Aug 21, 2020 at 12:16

Let $$A=\dfrac{a^4}{a^2+1}$$. $$A=\dfrac{a^4}{a^2+1}<\dfrac{a^4}{a^2}=a^2 \text{ and }b>A$$ Thus you need to show that $$b\notin (A,a^2)$$.

$$a^2$$ is an integer and $$a^2-A=a^2-\dfrac{a^4}{a^2+1}=\dfrac{a^2}{a^2+1}<1$$. Thus the interval $$(A,a^2)$$ cannot contain an integer and $$b\notin (A,a^2)$$. So $$b\geq a^2$$.

• I actually struggled to comprehend your solution. Probably next time I'm going to learn this. Thank you Aug 21, 2020 at 12:20

I finally understand how to. This is how I did it (since I already know how to solve, I didn't put words on it).

$$b > \frac {a^4}{a^2 +1}$$

$$\frac {a^4}{a^2 +1} = a^2 - 1 + \frac {1}{a^2 +1}$$

$$b > a^2 - 1 + \frac {1}{a^2 +1}$$

$$b > a^2 - 1 + \frac {1}{a^2 +1} > a^2 - 1$$

$$b > a^2 - 1$$

$$b \geq a^2 - 1 +1$$

$$b \geq a^2$$ .