<s>[INST] Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $. Show $xy(x^2+y^2) \leq 2$ 

Question: Two real numbers, $x$ and $y$, satisfy the condition $x + y = 2 $.
    Show $xy(x^2+y^2) \leq 2$



What I have attempted:
Consider $$x+y=2$$
$$ \Leftrightarrow (x+y)^2 = 2^2 $$
$$ \Leftrightarrow (x+y)^2 = 4 $$
Notice that $$ x^2 + y^2 = (x+y)^2 - 2xy $$
But $(x+y)^2 = 4$
$$ \Longrightarrow x^2 + y^2 = 4-2xy $$
Now we have this inequality; $$xy(x^2+y^2) \leq 2$$
$$\Leftrightarrow xy(4-2xy) \leq 2 $$
$$ \Leftrightarrow 2xy(2-xy) \leq 2 $$
$$ \Leftrightarrow 2(2xy-x^2y^2) \leq 2 $$
Now I am stuck, how should I continue also is there another way to approach this question?
 [/INST] Another way:
Let $x=r\cos \theta$; $y=r\sin\theta$. It is clear that we need only consider the case when $0 \le \theta \le \frac \pi 2$.
Since $x+y=2$ then $0\le r \le 2^\frac12$
(i.e., the longest distence from $(0,0)$ to the line $x+y=2$ when $0 \le \theta \le \frac \pi 2$).
Therefore, $$xy(x^2+y^2)=\frac12r^4\sin 2\theta \le \frac12 (2^\frac12)^4=2.$$
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