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

Question

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

I can prove this (I think) based on intuition, but not algebraically. My attempts as following.

Attempt 1 (intuition):

I have $(x+y)^2=x^2+2xy+y^2$. I take the RHS and write: $$ x^2+2xy+y^2=x^2+2xy+y^2 \tag 2 $$ In the RHS I discard the term $2xy$ and multiply with 2, so: $$ x^2+2xy+y^2=2(x^2+y^2) \tag 3 $$ And now I can (?) introduce an equality: $$(x+y)^2\leq 2(x^2+y^2) \tag 4 $$

Am I allowed to do this? Is it correct?

Attempt 2 (algebra):

Start with $(x+y)^2=x^2+2xy+y^2$.

Subtract $-2xy$ and add $x^2$ and $y^2$ on both sides:

\begin{align} (x+y)^2-2xy+x^2+y^2&=x^2+2xy+y^2-2xy+x^2+y^2 \tag 5\\\iff\\ (x+y)^2-2xy+x^2+y^2&=2x^2+2y^2 \tag 6 \end{align} Stuck, what have I missed here?

Answer 1

The standard way is

$$(x+y)^2\leq 2(x^2+y^2)\iff x^2+2xy+y^2\le 2x^2+2y^2$$$$\iff x^2-2xy+y^2=(x-y)^2 \ge 0$$

Answer 2

$$x^2+2xy+y^2\leq 2x^2+2y^2\;\; ?$$ $$ 0 \leq x^2-2xy+y^2 \;\;?$$ $$ 0\leq (x-y)^2 \;\;\checkmark$$