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> Let $a, b \in \mathbb{R}$ such that $n^{4a-\log\_5{n^2}} \leq 25^{40-b^2}$ for all positive real number $n$. Find the maximum value of $M=a^2+b^2+a-3b$.
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What I have tried:

Taking the logarithm base 5 of both sides from the inequality, we get $(4a-2\log\_5{n})\log\_5n\leq2(40-b^2)$

Now, let $x=\log\_5 n$, we get $x(4a-2x)\leq2(40-b^2)$, and this is equivalent to $x^2+2ax-b^2+40
\geq0$

The new inequality is true if and only if the discriminant $4a^2-4(40-b^2)\leq0$, this is equivalent to $a^2+b^2\leq40$, so we have $\max(a^2+b^2)=40$

This gave me hope of finding the maximum value of $M$ by finding the maximum value of $a-3b$ but I have no idea how to do it.

I'd like to call for some help. Thank you!