Find minima of the multivariable function $\frac{4}{x^2+y^2+1}+2 xy$ $$f(x,y)=\frac{4}{x^2+y^2+1}+2 xy \\ \text{within the  domain: }1/5\leq x^2+y^2\leq 4$$
I am able to find the maximum of the function at $x^2+y^2=4$ by substituting x,y for $\cos(t)$ and $\sin(t)$ and I am able to start working for a solution in the lower boundary: 
$$ x=\frac{1}{\sqrt5}\cos(t), \  y=\frac{1}{\sqrt5}\sin(t) \\ 
\frac{4}{\frac{6}{5}}+\frac{2}{\sqrt5}(\cos(t)\sin(t))\\ 
\frac{df}{dt}=\frac{2}{\sqrt5}\cos(2t)=0\\
\tan(t)=1\\
x=\frac{1}{\sqrt5}\lvert\cos(\frac{\pi}{4})\rvert \ ,  \ \ y=\frac{1}{\sqrt5}\lvert\sin(\frac{\pi}{4})\rvert \\
x=\pm\frac{1}{\sqrt{10}},y=\pm\frac{1}{\sqrt{10}}$$
When i resubstitute this into the original function I get the wrong answer. 
$$f\left(\frac{1}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)=\frac{4}{\frac{12}{10}}+\frac{2}{10}=53/15
$$
 A: $f_x(x,y) = 0 \to \dfrac{-8x}{(x^2+y^2+1)^2} = 2y$
$f_y(x,y) = 0 \to \dfrac{-8y}{(x^2+y^2+1)^2} = 2x$.
Thus $(x,y) = (0,0)$ is a critical point, also $\dfrac{x}{y} = \dfrac{y}{x} \to x = \pm y$
Consider $g(x) = f(x,x) = \dfrac{4}{2x^2+1} + 2x^2$ with $1/5 \leq 2x^2 \leq 4$. let $u = 2x^2$, then $h(u) = g(x,x) = \dfrac{4}{u+1} + u$, $1/5 \leq u \leq 4\to h'(u) = 1-\dfrac{4}{(u+1)^2} = 0 \iff u+1 = \pm 2 \to u =1,-3 \to u = 1$, since $u > 0$.
$h(1) = 3, h(1/5) = 53/15, h(4) = 24/5$.
And $r(x) = f(x,-x) = \dfrac{4}{2x^2+1} - 2x^2 = \dfrac{4}{u+1}-u = q(u)$, for $1/5 \leq u \leq 4$. 
$q'(u) = -1-\dfrac{4}{(u+1)^2} < 0 \to q(u) \geq q(4) = 4/5-4$.
Thus the min value of $f(x,y)$ is $4/5-4$.
A: You know that $1/5\leq x^2+y^2\leq 4$, so:
$$\frac{4}{x^2+y^2+1} \geq \frac{4}{4+1}=\frac{4}{5}$$
In addiction by AM-GM:
$$2|xy| \leq x^2+y^2 \leq 4$$
$$|xy| \leq 2$$
And: 
$$2xy \geq -4$$
So let's try to find such a $x,y$ that:
$$xy=-2,x^2+y^2=4$$
But for $x=-\sqrt{2}$ and $y=\sqrt{2}$ both equalities are satisfy, so minimum is:
$$\frac{4}{5}-4$$
