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
 A: 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.

A: 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,
$$
\begin{align}
x+y
&=\frac{x^3+y^3}{x^2-xy+y^2}\\
&\le\frac4{x^2+y^2}\\[6pt]
&\to0
\end{align}
$$
A: 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 $(x,y) = \{(\sqrt[3]{2},0), (0,\sqrt[3]{2})\}$
