<s>[INST] then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
 [/INST] Note that
Since
$$
3x^2+3y^2y'=0
$$
we have
$$
y'=-\frac{x^2}{y^2}
$$
Therefore,
$$
\begin{align}
0
&=(x+y)'\\[6pt]
&=1+y'\\
&=1-\frac{x^2}{y^2}\\
\end{align}
$$
At $x=y=1$, we get a maximum of $2$.
$x=-y$ doesn't happen, but $\frac xy\to-1$ as $x\to\pm\infty$. Since $xy\le\frac{x^2+y^2}2$, we have $x^2-xy+y^2\ge\frac{x^2+y^2}2$. Therefore,
$$
\begin{align}
x+y
&=\frac{x^3+y^3}{x^2-xy+y^2}\\
&\le\frac4{x^2+y^2}\\[6pt]
&\to0
\end{align}
$$
</s>