$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$ I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.

Given the function 
$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$
Prove that the taylor polynomial of degree $4$ of $f$ is equal to
  $5+x^4+y^4$.

First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $
$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$
Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$ 
Therefore: $f(x,y)=x^4+y^4+5+4R_3$
I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.

Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$

New edit 
I understand this now thanks to hint of Hagen von Eitzen, thanks !
The third question is:

Determine what kind of stationary point you have in $(0,0)$.


Write $x^4=2x^4-x^4$ and similarly $y^4=2y^4-y^4$
$$(x^2-y^2)^2 \ge 0$$
$$x^4+y^4-2x^2y^2 \ge 0$$
$$2x^4-x^4+2y^4-y^4-2x^2y^2 \ge0$$
$$2x^4+2y^4 \ge x^4+y^4+2x^2y^2$$
$$x^4+y^4 \ge \dfrac{(x^2+y^2)^2}{2}$$
