<s>[INST] Prove $\frac{b^{4} - 24 a^{4}}{ab} \ge 3b^{2} + 6ab - 28a^{2} $ I made up a simple problem. I wonder if there are other ways to solve this problem? or maybe whether or not the inequality seems very obvious?

If $a,b$ are positive real numbers and $b$ is at least twice of $a$, prove that 
$$ \frac{b^{4} - 24 a^{4}}{ab} \ge 3b^{2} + 6ab - 28a^{2} $$
Solution:
We want to prove
$$ b^{4} - 3ab^{3} - 6a^{2}b^{2} + 28a^{3}b - 24a^{4} \ge 0 $$
Now let $b = at$ with $t \ge 2$. So the inequality becomes:
$$ t^{4} - 3t^{3} - 6t^{2} + 28t - 24 \ge 0 $$
Notice equality when $t=2$, so $(t-2)$ is a factor:
$$ (t-2)(t^{3} - t^{2} -8t + 12)  \ge 0 $$
$$ (t-2)(t-2)(t^{2}+t-6) = (t-2)^{3}(t+3)  \ge 0 $$
 [/INST] Your inequality equivalent to
$b^4 -24a^4 \geq ab(3b^2 +6ab-28a^2)$
We have: $LHS-RHS = (b-2a)^3 (3a+b) \geq 0$ (because: $b$  is at least twice of $a$)
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