If $ 5x+12y=60$ , what is the minimum of $\sqrt{x^2+y^2}$?

Note that $$(5x+12y)^2+(12x-5y)^2=(13^2)(x^2+y^2).$$ Since $5x+12y$ is given, we minimize $x^2+y^2$ by minimizing $(12x-5y)^2$. The minimum value of this is $0$. It follows that the minimum value of $\sqrt{x^2+y^2}$ is $\dfrac{60}{13}$.

Hint: Consider the graph of the line $5x+12y = 60$.

Hint: The minimum value occurs when the circle with origin as center, is tangential to the above line.

Hint: Find the point of tangency. Use the fact that a tangent is perpendicular to the radius.

Apply (weighted) QM-AM, we have

$\sqrt{\frac{x^2 + y^2}{169}} = \sqrt{\frac{25 \times \frac{x^2}{25} + 144 \times \frac{y^2}{144}}{169}} \geq \frac{ 25 \times \frac{x}{5} + 144 \times \frac{y}{12} } { 169 } = \frac{ 5x+12y}{169} = \frac{60}{169}$.

SO $\sqrt{x^2+y^2} \geq \frac{60}{13} $