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...
 A: 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*}
A: 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 $b>\frac{a^{4}}{a^{2}+1}$
A: 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 $b\geq a^2$.
A: 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).
$$ b > \frac {a^4}{a^2 +1}$$
$$\frac {a^4}{a^2 +1} = a^2 - 1 + \frac {1}{a^2 +1}$$
$$b >  a^2 - 1 + \frac {1}{a^2 +1}$$
$$b >  a^2 - 1 + \frac {1}{a^2 +1} > a^2 - 1$$
$$b > a^2 - 1$$
$$b \geq a^2 - 1 +1$$
$$b \geq a^2$$ .
