<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] Hint: Consider $f(x)= x(2-x)(x^2+(2-x)^2)$. Then $f'(x)=-8 (x-1)^3$.
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