Find the minimum value of the given function by using trigonometry
$$(x_1-x_2)^2 + \left(\dfrac{9}{x_1} – \sqrt{1-x_2^2}\right)^2$$
I know the distance formula method but is there any other suitable method to solve this? I’m thinking along the lines of trigonometry or AM-GM inequality.
Best Answer
Expand as usual, and using Cauchy-Schwarz inequality: $f(x_1,x_2) = x_1^2 - 2x_1x_2 + \dfrac{81}{x_1^2} - \dfrac{18\sqrt{1-x_2^2}}{x_1}+1 = 1+x_1^2+\dfrac{81}{x_1^2} - 2\left(x_1x_2+\dfrac{9\sqrt{1-x_2^2}}{x_1}\right)\ge 1+x_1^2+\dfrac{81}{x_1^2} - 2\sqrt{x_1^2+\dfrac{81}{x_1^2}}\sqrt{x_2^2+(1-x_2^2)} = 1 + y^2- 2y = g(y)$, with $y = \sqrt{x_1^2+\dfrac{81}{x_1^2}}\ge 3\sqrt{2}$ by AM-GM inequality. Thus the problem reduces to finding the min value of $g(y) = y^2-2y+1, y \ge 3\sqrt{2}$. We have again: $g(y) = (y-1)^2 \ge (3\sqrt{2}-1)^2= 19-6\sqrt{2}$, and thus $f(x_1,x_2)_{\text{min}} = 19-6\sqrt{2}$ and this min value achieved when $x_1 = \pm 3, \dfrac{x_1}{x_2}=\dfrac{\dfrac{9}{x_1}}{\sqrt{1-x_2^2}}\implies x_1 = \pm 3, x_2 = \sqrt{1-x_2^2}\implies x_2^2=1-x_2^2\implies x_2 = \pm \dfrac{1}{\sqrt{2}}$.