<s>[INST] Supremum of $x^2y^2(x^2+y^2)$ Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is

*

*$3n^2$

*$2n^6$

*$4n^6$

*$n^6$
My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express $f(x, y) $ as a function of $x$ by putting $y=2n-x$. Then applying calculus we can find the maximum for some value of $x$.
But calculation will be cumbersome. Is there any smart way?
Thanks in advance.
 [/INST] By the AM-GM inequality, we have
$$xy \leqslant \left(\frac{x+y}{2}\right) ^2 =n^2,$$
and
$$xy(x^2+y^2) \leqslant \frac{1}{2} \cdot \left(\frac{2xy+(x^2+y^2)}{2}\right)^2=\frac{1}{2} \cdot \left(\frac{(x+y)^2}{2}\right)^2 = 2n^4.$$
Therefore
$$x^2y^2(x^2+y^2) \leqslant 2n^6.$$
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