How to find the minimum of $\dfrac{|3a+4b-1|}{\sqrt {a^2+b^2}}$ if $\dfrac{1}{a^2}+\dfrac{1}{b^2}=25$ I want to find the minimal value of $\dfrac{|3a+4b-1|}{\sqrt {a^2+b^2}}$ with $\dfrac{1}{a^2}+\dfrac{1}{b^2}=25$
The answer is $\dfrac{12}{5}$
My attempt was:$\dfrac{1}{a^2}+\dfrac{1}{b^2}=25\implies a^2+b^2=25a^2b^2\implies \sqrt{a^2+b^2}=5|a||b|$
Thus, I turn the original formula into $$\frac{|3a+4b-1|}{5|a||b|}=\dfrac{1}{5}\frac{|3a+4b-1|}{|a||b|}$$
That is I only need to find the minimum of $\dfrac{|3a+4b-1|}{|a||b|}$, but I got stuck here due to absolute values.
I also tried to use the relation of $\dfrac{1}{a^2}+\dfrac{1}{b^2}=25$ to make $a=...b$ and tried to make the original formula to a function only involves $a$ (or $b$) and take the derivative, but this seems won't work so well, and the calculation is terrifying.
By observation, I think this question might be able to deal with by geometry constructions, and I also tried this, even though I failed finally. But if it's possible, I really prefer only algebra way to solve this problem. Any help on this? Thanks in advance.

We'll prove that the minimal value is $0$, for which it's enough to prove that the system
$3a+4b=1$ and $\frac{1}{a^2}+\frac{1}{b^2}=25$ has real solutions and we need to show that the  equation
$$\frac{1}{a^2}+\frac{16}{(1-3a)^2}=25$$
has real root.
Let $f(a)=\frac{1}{a^2}+\frac{16}{(1-3a)^2}-25.$
Thus, $f\left(\frac{1}{2}\right)>0$ and $f(1)<0$, which says that on $(0.5,1)$ there is a real root.
If we say about positives $a$ and $b$, so for $a=\frac{1}{3}$ and $b=\frac{1}{4}$ we'll get a value $\frac{12}{5}$.
We'll prove that it's a minimal value.
Indeed, since easy to show that $3a+4b>1,$ it's enough to prove that $$\frac{3a+4b-1}{\sqrt{a^2+b^2}}\geq\frac{12}{5}$$ or
$$3a+4b\geq1+12ab$$ or
$$\frac{5ab(3a+4b)}{\sqrt{a^2+b^2}}\geq\frac{25a^2b^2}{a^2+b^2}+12ab$$ or
$$5(3a+4b)\sqrt{a^2+b^2}\geq12a^2+25ab+12b^2$$ or
$$5\sqrt{a^2+b^2}\geq4a+3b,$$ which is true by C-S:
$$5\sqrt{a^2+b^2}=\sqrt{(4^2+3^2)(a^2+b^2)}\geq4a+3b.$$
