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.
Best Answer
Let $x := 1/a, y := 1/b$. Then, $\dfrac{|3a+4b-1|}{|ab|} = |3x + 4y - xy|$ and hence we need to consider $|3x + 4y - xy|$ under the condition that $x^2 + y^2 = 1$.
We can use the Lagrange multiplier theorem or simply let $(x,y) = (\cos\theta,\sin\theta)$ and consider $|3\cos\theta + 4\sin\theta - \cos\theta\sin\theta|$.