Find the minimum value of the function $f(x, y)$ for $x, y > 0:$
$f(x,y)=\sqrt{x^{2}+1}+\sqrt{\frac{y^{2}}{x^{2}}+4}+\sqrt{\frac{9}{y^{2}}+9}$
Solution
Let, $a = x,b = \frac{y}{x},c = \frac{3}{y}$
Then $a,b,c > 0$ and $abc = x.\frac{y}{x}.\frac{3}{y} = 3$
So $f = \sqrt {{a^2} + 1} + \sqrt {{b^2} + 4} + \sqrt {{c^2} + 9} $
Now, Minkowski inequality (in its 2-dimensional form)
for ${a_i},{b_i} \ge 0$, $\sum\limits_{i = 1}^n {\sqrt {a_i^2 + b_i^2} } \ge \sqrt {{{\left( {\sum\limits_{i = 1}^n {{a_i}} } \right)}^2} + {{\left( {\sum\limits_{i = 1}^n {{b_i}} } \right)}^2}} $
$\therefore f = \sqrt {{a^2} + {1^2}} + \sqrt {{b^2} + {2^2}} + \sqrt {{c^2} + {3^2}} \ge \sqrt {{{(a + b + c)}^2} + {{(1 + 2 + 3)}^2}} $
$\therefore f \ge \sqrt {{{(a + b + c)}^2} + 36} $
Now, $A.M. \ge G.M$ or $\frac{{a + b + c}}{3} \ge {(abc)^{1/3}}$
$ \Rightarrow a + b + c \ge {3^{4/3}}$
$\therefore f \ge \sqrt {{3^{8/3}} + 36} $
$ \Rightarrow f \ge \sqrt {{{9.9}^{1/3}} + 9.4} $
$ \Rightarrow f \ge 3\sqrt {{9^{1/3}} + 4} $
$\therefore{f_{\min }} = 3\sqrt {{9^{1/3}} + 4} $