Not only AM-GM 2 (The Other AM-GM Part 4)

Similar solution with Fidel Simanjuntak 's

Using Cauchy-Schwarz inequality as follows:

$\begin{aligned} (5x+12y)^2 & \le (5^2+12^2) (x^2+y^2) & \small \color{#3D99F6} \text{Square root both sides} \\ \color{#3D99F6} 5x+12y & \le 13 \sqrt {x^2+y^2} & \small \color{#3D99F6} \text{Note that }5x+12y = 60 \\ \implies \sqrt {x^2+y^2} & \ge \frac {60}{13} \end{aligned}$

$\implies a - b = 60-13 = \boxed{47}$

Relevant wiki: 2D Coordinate Geometry - Problem Solving

Given , $\displaystyle \dfrac{x}{12}+\dfrac{y}{5}=1$

We need to find the Radius of a circle $(R)$ centered at origin $(0,0)$ which just touches the line at P. Let the equation of the circle be $x^2+y^2=R^2$

So the distance of $\displaystyle \dfrac{x}{12}+\dfrac{y}{5}=1$ from $(0,0)$ is :

$\displaystyle \sqrt{x^2+y^2}_{min}=R_{min}=\dfrac{|12\times 0+5\times 0-60|}{\sqrt{12^2+5^2}}=\boxed{\dfrac{60}{13}}$

making the answer $\boxed{60-13=47}$

Relevant Wiki - Cauchy-Schwarz Inequality

By Cauchy-Schwarz Inequality, we have

$(5x+12y)^2 \leq (5^2 + 12^2)(x^2 + y^2)$

$60^2 \leq 169(x^2+y^2)$

$\dfrac{3600}{169} \leq (x^2 + y^2)$

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

Hence, $a-b = 47$ .