Find extreme values of $f(x,y) = x^2 + y^2$ subject to $g(x,y) = xy = 1$
Quite easy calculations, but I get that values of extreme points $f(1,1) = 2$ and $f(-1,-1) = 2$. Still it is easy to determine that they are minimums, because of the positive paraboloid, but I should have showed it without using Hessian. Is there some way to do it?
On
Notice that $$ f(x,y)-f(1,1) = x^2+y^2-2 \overset{(*)}{=} x^2+\frac{1}{x^2}-2 =\left(x-\frac{1}{x}\right)^2 \geq 0, \tag{1} $$ where I used the constraint $xy=1$ in the equality $(*)$. It follows from $(1)$ that $f$ has a minimum at $(1,1)$. A similar calculation shows that $f$ also has a minimum at $(-1,-1)$.
One possible way to approach:
\begin{align*} f(x,y) = x^{2} + y^{2} = x^{2} + \frac{1}{x^{2}} & = \left(x - \frac{1}{x}\right)^{2} + 2 \geq 2 \end{align*}
The same result can be obtained using the AM-GM inequality:
\begin{align*} f(x,y) = x^{2} + y^{2} \geq 2\sqrt{x^{2}y^{2}} = 2|xy| = 2xy = 2 \end{align*}
Such value is attained when
\begin{align*} \left(x - \frac{1}{x}\right)^{2} = 0 & \Longleftrightarrow x^{2} - 1 = 0 \Longleftrightarrow x = \pm 1 \end{align*}
That is to say, the critical points are $(1,1)$ and $(-1,-1)$.
Hopefully this helps!