<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] If $x,y$ are real numbers then there is no absolute minimum. In fact the value of $x+y$ can be made as close to zero as we like. We have $y = \sqrt[3]{2 - x^3}$
Obviously for very large $x$ we have that $y \approx -x$. Additionally the value must be positive, as $y > -x$, from the condition.
On the other side if we add the condition that $x,y$ are positive, then we have:
$$(x+y)^3 \ge x^3 + y^3 = 2 \implies x+y \ge \sqrt[3]{2}$$
And the minimum is obtained for $(x,y) = \{(\sqrt[3]{2},0), (0,\sqrt[3]{2})\}$
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