How to prove $x^4+y^4\leq (x^2+y^2)^2$?

Question

I want to prove: $$x^4+y^4\leq (x^2+y^2)^2 \tag 1$$

We always have $x^2\geq 0$ and $y^2\geq 0$. Therefore we also have $x^4\geq 0$ and $y^4\geq 0$.

Addition of the inequalities gives $$ x^4+y^4\geq 0 \tag 2 $$ I now add $2x^2y^2$ to both sides, so \begin{align} x^4+y^4+2x^2y^2&\geq 2x^2y^2 \tag 3 \\ (x^2+y^2)^2&\geq 2x^2y^2 \tag 4 \end{align} I'm stuck here...

Answer 1

Hint: $$2x^2y^2\geq 0$$ for all real $x,y$ $$(x^2+y^2)^2=x^4+y^4+2x^2y^2$$

Answer 2

$$(x^2+y^2)^2=x^4+2x^2y^2+y^4\geqslant x^4+y^4.$$