<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] You have $y=2-x$, so 
$$
x^2+y^2=(x+y)^2-2xy=4-2xy=4-2x(2-x).
$$
Then
$$
xy(x^2+y^2)=x(2-x)(4-2x(2-x))
=2r(2-r),
$$
where $r=x(2-x)$. 
So consider the parabola $-2r^2+4r$. It's maximum occurs at $r=-4/-4=1$ ("$-b/2a$"), so 
$$
2r(2-r)\leq 2(2-1)=2.
$$
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