the minmum value of #9

Let's take a LaGrange Multipliers approach here. Taking $f(x,y) = (x+\frac{1}{x})^2 + (y+\frac{1}{y})^2 - \frac{1}{2}$ and $g(x,y)=x+y=1$ , we compute $\nabla f = \lambda \cdot \nabla g$ . This now yields:

$2(x+\frac{1}{x})(1-\frac{1}{x^2}) = \lambda;$ (i)

$2(y+\frac{1}{y})(1-\frac{1}{y^2}) = \lambda;$ (ii)

and equating (i) with (ii) gives:

$\frac{x^4-1}{x^3} = \frac{y^4-1}{y^3};$

or $x^4y^3-x^3y^4 = y^3-x^3$ ;

or $x^3y^3(x-y) = -(x-y)(x^2+xy+y^2)$ ;

or $x=y$ (iii); $(xy)^3 = -(x^2+xy+y^2)$ (iv)

Since (iv) is a contradiction for all $x,y \in \mathbb{R^{+}}$ we only admit (iii) $\Rightarrow x = y = \frac{1}{2}.$ Taking the Hessian matrix of $f$ yields:

$F(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 1+\frac{3}{x^4} & 0 \\ 0 & 1 + \frac{3}{y^4} \end{bmatrix}$

and evaluated at the critical point $x = y = \frac{1}{2}$ :

$F(1/2,1/2) = \begin{bmatrix} 49 & 0 \\ 0 & 49 \end{bmatrix} = 49 \cdot I_{2x2}$ (which is positive-definite and $(1/2,1/2) \Rightarrow$ a global minimum over $x,y \in \mathbb{R^{+}}$ ).

Thus the minimum value computes to $f(1/2,1/2) = 2(\frac{1}{2} + 2)^2 - \frac{1}{2} = \boxed{12}.$