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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

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    You could try to judge by symmetry. – 2017-01-12

3 Answers 3

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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.

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    (+1) Sorry, I didn't see your answer when I was writing mine. Even so, they look quite different although they are essentially the same. – 2017-01-12
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    @robjohn: No problem. Actually the first half of Stefan's answer is also the same (only in condensed form) and appeared only shortly before I was finished writing mine. – 2017-01-12
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    Indeed, I just noticed that. Looking at the end of his answer, I thought he was finding a minimum, and I was about to write a comment. Then I read the beginning more closely. The only answer I saw before I posted was Dr. Sonnhard Graubner's. – 2017-01-12
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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, $$ \begin{align} x+y &=\frac{x^3+y^3}{x^2-xy+y^2}\\ &\le\frac4{x^2+y^2}\\[6pt] &\to0 \end{align} $$

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    Could I ask why $x=-y$ doesn't happen, while $\frac{x}{y} \rightarrow -1$ does as $x \rightarrow \pm \infty$? – 2017-01-13
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    OK, $x=-y$ doesn't happen because of the constraint. I don't get the second part. – 2017-01-13
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    Which second part? $\frac xy\to-1$? $x^2-xy+y^2\ge\frac{x^2+y^2}2$? $x+y\le\frac4{x^2+y^2}$? – 2017-01-13
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    I meant the second part of my question, so $\frac{x}{y} \rightarrow -1$. But I figured it out now. Thanks – 2017-01-13
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    Oh, that makes sense. Sorry, it was late and I was tired. Glad you've got it all now. – 2017-01-13
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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 $(x,y) = \{(\sqrt[3]{2},0), (0,\sqrt[3]{2})\}$