Inspired By Sharky Kesa

Relevant wiki: Classical Inequalities - Problem Solving - Intermediate

Note that $16x^2+\frac{1}{x^2}+y^2+\frac{y}{2x}=16x^2+\frac{1}{x^2}+\left(y+\frac{1}{4x}\right)^2-\frac{1}{16x^2} \ge 16x^2+\frac{1}{x^2}-\frac{1}{16x^2}$

Now $16x^2+\frac{1}{x^2}-\frac{1}{16x^2} =16x^2+\frac{15}{16x^2} \ge 2 \sqrt{15} (\because \text{AM-GM} )$

So $m$ is $2^2 \times 15=60$ . The equality is true when $y=-\frac{1}{4x}$ and $16x^2=\frac{15}{16x^2}$ .

Solving this system of equations gives us $x=\pm \sqrt[4]{\frac{15}{256}}$ and $y= \mp \sqrt[4]{\frac{1}{15}}$ .

$\boxed{1.-}$

(Calculus solution 1)

Let $f : A = \{(x,y) \in \mathbb{R}^2; x \neq 0 \wedge y \neq 0\} \to \mathbb{R}$ be the differentiable function defined as $f(x, y) = 16x^2 + y^2 + \frac{1}{x^2} + \frac{y}{2x}$ in the open set $A$ . A global extremum is always a local extremum. Due to $A$ is an open set and $f$ is a differentiable function in $A$ , I'm going to look for local extremums, being a ( a or one doesn't mean the only one) necessary condition for finding them: $\frac{\partial f}{\partial x} = 32x - \frac{2}{x^3} - \frac{y}{2x^2} = 0, (1)$ $\frac{\partial f}{\partial y} = 2y + \frac{1}{2x} = 0 \Rightarrow y = \frac{-1}{4x}, (2)$ Substituing $y$ at (1) from (2) we get: $256x^4 -16 + 1 = 0 \Rightarrow x^4 = \frac{15}{256} \Rightarrow$ The only 2 real solutions for $x$ are $x = \pm \sqrt[4]{\frac{15}{256}} \Rightarrow y = \frac{-1}{4x}$ . Making the hessian matrix at these two points(which is defined possitive), we make sure (it's sufficient) that at these points are local minimums, and because of a global extremum is a local extremum the only possibilities(in this case) for attaining a global extremum are these two points. Substituing at $f(x,y)$ at these 2 points, we get the minimum of this function what is $\sqrt{60}$ .

$\boxed{2.-}$

(Calculus solution 2)

It's clear by looking the above expression that for attaining a minimum of this function, $y$ has to have different sign that $x$ ...

Make $y = \frac{1}{t}$

$16x^2 + y^2 + \frac{1}{x^2} + \frac{y}{2x} = 16x^2 + \frac{1}{t^2} + \frac{1}{x^2} + \frac{1}{2xt}$ Let's make now $t = -4x$ and come back to the original equation: $16x^2 + y^2 + \frac{1}{x^2} + \frac{y}{2x} = 16x^2 + \frac{1}{16x^2} + \frac{1}{x^2} - \frac{1}{8x^2} = f(x)$ The rest is to minimize $f(x)$ ...

For it you can calculate $f'(x)$ and $f''(x)$ or use Wolfram alpha, or ... .

Let $f(y) = y^2 + \frac{1}{2x}y+(16x^2 +\frac{1}{x^2})$ be the function. By completing the square, we have $f(y) = (y+\frac{1}{4x})^2 +(16x^2 +\frac{15}{16x^2})$ Notice that the minimum of $f(y)$ is when $y = -\frac{1}{4x}$ . Thus, we have the minimum of f(y), which is $16x^2 + \frac{15}{16x^2}$ . By AM-GM, we have $16x^2 + \frac{15}{16x^2}\geq 2\sqrt{16x^2 \times \frac{15}{16x^2}} = 2\sqrt{15}=\sqrt{60}$ - Equality holds when $16x^2 = \frac{15}{16x^2} \Rightarrow x = \pm \sqrt[4]{\frac{15}{256}}$ Hence, the minimum value of $f(y)$ is $\sqrt{60}$ , when $x=\pm \sqrt[4]{\frac{15}{256}}$ and $y = \mp\sqrt[4]{\frac{1}{15}}$