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

Question

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...

Answer 1

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

Answer 2

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*}

Answer 3

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 $b\geq a^2$.

Answer 4

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).

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

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

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

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

$$b > a^2 - 1$$

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

$$b \geq a^2$$ .