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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...

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  • $\begingroup$ Thank you @MPW for editing. $\endgroup$ Aug 21, 2020 at 3:59
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    $\begingroup$ How do you get $b>a^2?$ For example, $a=b=1$ satisfies the first inequality. $\endgroup$ Aug 21, 2020 at 3:59
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    $\begingroup$ if you take a=b=1 don't you have b=a^2 and your inequality is satisfied? $\endgroup$
    – PAM1499
    Aug 21, 2020 at 4:00
  • $\begingroup$ $-1+\frac {1} {a^2 +1}$ is a small difference like it won't affect after all (by some brute force). $\endgroup$ Aug 21, 2020 at 4:03
  • $\begingroup$ Since I want an elegant solution. I want to do it in casework, but how?? $\endgroup$ Aug 21, 2020 at 4:05

4 Answers 4

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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*}

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  • $\begingroup$ 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. $\endgroup$ Aug 21, 2020 at 12:13
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    $\begingroup$ @hansduran0123: Yes, your answer is fine. $\endgroup$
    – quasi
    Aug 21, 2020 at 21:46
  • $\begingroup$ Yes. Thanks to you. $\endgroup$ Aug 22, 2020 at 10:54
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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 $b>\frac{a^{4}}{a^{2}+1}$

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    $\begingroup$ The negation of $b>a^2$ is $b\le a^2$ $\endgroup$
    – egreg
    Aug 21, 2020 at 12:14
  • $\begingroup$ 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. $\endgroup$ Aug 21, 2020 at 12:16
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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 $b\geq a^2$.

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

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

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

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

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

$$b > a^2 - 1$$

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

$$b \geq a^2$$ .

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