Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches) 
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.

I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\min(\sqrt{x^2+y^2})=3$.
I'm wondering is it possible to solve this problem differently?

By Cauchy-Schwarz inequality,
$\sqrt{x^2+y^2}\ge\frac{15}{\sqrt{3^2+4^2}}=3$
and by the case of Cauchy-Schwarz equality, this value is attained, hence it is the min you look for.
