<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] From 
$$
 2 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = (x+y) \frac{(x-y)^2+x^2+y^2}2
$$
it follows that $x+y> 0$. On the other hand, for arbitrary $t > 0$
$$
 x =-t \quad , \quad y = \sqrt[3]{2+t^3}
$$
satisfies $x^3+y^3 =2$ and
$$
 x + y = \sqrt[3]{2+t^3} - t = \frac{2}{(\sqrt[3]{2+t^3})^2 + t \sqrt[3]{2+t^3} + t^2} \to 0
$$
for $t \to \infty$. Therefore the infimum is zero and a minimum
does not exist.

</s>