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What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.

I plotted the graph of $x^3+y^3+xy=1$. It seems the minima occurs when $x=y$ but I don't know why that will be true. Any ideas?

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    Please see here (asked only an hour ago): http://math.stackexchange.com/questions/2107048/find-a-minimum-of-x2y2-under-the-condition-x33xyy3-1 – 2017-01-21
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    use the Lagrange Multiplier function to find the searched minimum – 2017-01-21
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    are the variables positive reals? – 2017-01-21
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    @Dr.SonnhardGraubner They are reals. Maybe positive or negative. – 2017-01-21
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    @Rohan No, that is slightly different. It has $3xy$ term while mine has $xy$ term only. – 2017-01-21
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    Yes yes that's what even the OP of that question first wrote. You may not able to see a deleted answer which gave a full answer to your question. – 2017-01-21
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    @Rohan Can you copy paste the deleted answer here? I can't see it. – 2017-01-21
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    Unfortunately he didn't prove why $x=y $ satisfies the condition. – 2017-01-21
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    Why the downvote ? :( – 2017-01-21

2 Answers 2

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Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.

Hence, $x^3+y^3+xy=1$ gives $2u(4u^2-3v^2)+v^2=1$ or $v^2=\frac{8u^3-1}{6u-1}$.

But $u^2\geq v^2$.

Hence, $u^2\geq\frac{8u^3-1}{6u-1}$ or $\frac{2u^3+u^2-1}{6u-1}\leq0,$ which gives $\frac{1}{6}

where $u_1$ is a real root of the equation $2u^3+u^2-1=0$.

Hence, $x^2+y^2=4u^2-2v^2=4u^2-\frac{2(8u^3-1)}{6u-1}=\frac{2(4u^3-2u^2+1)}{6u-1}$, which is decreasing on $\left(\frac{1}{6},u_1\right]$, which says that the answer is $\frac{2(4u_1^3-2u_1^2+1)}{6u_1-1}.$

We can get $u_1$ by the Cardano's formula.

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    i think this is not true Michael – 2017-01-21
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    in the original post we have $$x^3+y^3+xy=1$$ – 2017-01-21
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    @Dr. Sonnhard Graubner Thank you! I fixed my post. – 2017-01-21
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The cubic $x^3+y^3+xy+1$ is symmetric in $y=x$. So lets rotate it by $\frac{\pi}{4}$. Let $x=\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}$ and $y=\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}$. Note that the function to minimize becomes $x`^2+y`^2$.

$$\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}+\frac{y`}{\sqrt{2}}\right)\left(\frac{x`}{\sqrt{2}}-\frac{y`}{\sqrt{2}}\right)-1=0$$

$$\frac{3x`y`^2}{\sqrt{2}}+\frac{x`^3}{\sqrt{2}}-\frac{y`^2}{2}+\frac{x`^2}{2}-1=0$$

We can solve for $y`^2$.

$$y`^2=\frac{-\sqrt{2}x`^3-x`^2+2}{3\sqrt{2}x`-1}$$

As this is positive solutions will only exist for $-\sqrt{2}x`^3-x`^2+2\ge0$. Note the root of $-\sqrt{2}x`^3-x`^2+2=0$ can be found by solving: $-2\left(2\left(\frac{x`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}\right)^2-1\right)=0$. Also note that $y`^2$ is strictly decreasing on this domain.

Using calculus to find the minimum of $x`^2+y`^2$ gives a value outside this domain so the minimum must be at the end of the domain, i.e. at the real root of $2\left(\frac{x`}{\sqrt{2}}\right)^3+\left(\frac{x`}{\sqrt{2}}\right)^2-1=0$. And the corresponding $y`$ is $0$. So rotating back to $x$ and $y$ shows that the minimum is on the line $x=y$ and is the real root of $2x^3+x^2-1$.